RD SHARMA SOLUTION CHAPTER-8 Transformation Formulae ICLASS 11TH MATHEMATICS-EDUGROWN

Chapter 8 Transformation Formulae Exercise Ex. 8.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

prove that ten 20o tan 30o tan 40o tan 80o = 1Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 5(viii)

Solution 5(viii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8

Solution 8

Chapter 8 Transformation Formulae Exercise Ex. 8.2

Question 3(i)

cos 55 to the power of 0 plus cos 65 to the power of 0 plus cos 175 to the power of 0 space equals space 0

Solution 3(i)

Question 18

I f space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses comma space P r o v e space t h a t space x y plus y z plus z x equals 0

Solution 18

G i v e n space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses equals k left parenthesis s a y right parenthesis
x equals fraction numerator k over denominator cos theta end fraction
y equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction
z equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction
x y plus y z plus z x equals k squared open square brackets fraction numerator 1 over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses cos theta end fraction close square brackets
equals k squared open square brackets fraction numerator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses plus cos theta plus cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator cos theta cos fraction numerator 4 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 4 straight pi over denominator 3 end fraction plus cos theta plus cos theta cos fraction numerator 2 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 2 straight pi over denominator 3 end fraction over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses begin display style fraction numerator minus square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals k squared open square brackets fraction numerator minus cos theta plus sin theta open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets
equals 0

H e n c e space P r o v e d

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

text LHS = end text sin 105 to the power of ring operator plus cos 105 to the power of ring operator
equals sin 105 to the power of ring operator plus cos left parenthesis 90 to the power of ring operator plus 15 to the power of ring operator right parenthesis
equals sin 105 to the power of ring operator minus sin 15 to the power of ring operator
equals 2 sin open parentheses fraction numerator 105 to the power of ring operator minus 15 to the power of ring operator over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 105 to the power of ring operator plus 15 to the power of ring operator over denominator 2 end fraction close parentheses
equals 2 sin 45 to the power of ring operator cos 60 to the power of ring operator
equals 2 fraction numerator 1 over denominator square root of 2 end fraction 1 half
equals fraction numerator 1 over denominator square root of 2 end fraction
equals cos 45 to the power of ring operator

Question 2(vi)

Solution 2(vi)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 6(v)

Solution 6(v)

Question 6(vi)

Solution 6(vi)

Question 6(vii)

Prove that:

cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction equals sin 7 θsin 8 straight theta

Solution 6(vii)

table attributes columnalign left end attributes row cell Consider text   the   left   hand   side   of   the   given   expression : end text end cell row cell straight L. straight H. straight S. equals cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction end cell row cell We text   know   that   2 cosAcosB = cos end text left parenthesis text A + B end text right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis end cell row cell Thus comma end cell row cell straight L. straight H. straight S. equals 1 half open square brackets text cos end text open parentheses straight theta text + end text straight theta over 2 close parentheses plus cos open parentheses straight theta minus straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses 3 straight theta text + end text fraction numerator 9 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses 3 straight theta minus fraction numerator 9 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses negative fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets text   end text text              end text left square bracket because cos left parenthesis negative straight theta right parenthesis equals cosθ right square bracket end cell row cell text           end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses minus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text           end text equals 1 half open square brackets cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell Also text   we   know   that , end text end cell row cell text cosD end text minus cosC equals 2 sin fraction numerator straight C plus straight D over denominator 2 end fraction sin fraction numerator straight C minus straight D over denominator 2 end fraction end cell row cell Therefore comma end cell row cell straight L. straight H. straight S. equals 1 half cross times 2 sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction plus straight theta over 2 over denominator 2 end fraction sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction minus straight theta over 2 over denominator 2 end fraction end cell row cell text         end text equals sin fraction numerator fraction numerator 16 straight theta over denominator 2 end fraction over denominator 2 end fraction sin fraction numerator fraction numerator 14 straight theta over denominator 2 end fraction over denominator 2 end fraction end cell row cell text        end text equals sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction end cell row cell text        end text equals straight R. straight H. straight S. end cell row blank row cell asterisk times Note colon text   Question   given   in   the   book   is   incorrect. end text end cell row cell text R. H. S .  should   be   equal   to   end text sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction. end cell end table

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 7(iv)

Solution 7(iv)

Question 7(v)

Solution 7(v)

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 8(v)

Solution 8(v)

Question 8(vi)

Solution 8(vi)

Question 8(vii)

Solution 8(vii)

Question 8(viii)

Solution 8(viii)

text LHS end text equals fraction numerator s i n 3 A cos 4 A minus sin A cos 2 A over denominator sin 4 A sin A plus cos 6 A cos A end fraction
equals fraction numerator 2 left parenthesis s i n 3 A cos 4 A minus sin A cos 2 A right parenthesis over denominator 2 left parenthesis sin 4 A sin A plus cos 6 A cos A right parenthesis end fraction
equals fraction numerator 2 s i n 3 A cos 4 A minus 2 sin A cos 2 A over denominator 2 sin 4 A sin A plus 2 cos 6 A cos A end fraction
equals fraction numerator s i n left parenthesis 4 A plus 3 A right parenthesis minus s i n left parenthesis 4 A minus 3 A right parenthesis minus open square brackets s i n left parenthesis 2 A plus A right parenthesis minus s i n left parenthesis 2 A minus A right parenthesis close square brackets over denominator cos left parenthesis 4 A minus A right parenthesis minus cos left parenthesis 4 A plus A right parenthesis plus cos left parenthesis 6 A plus A right parenthesis plus cos left parenthesis 6 A minus A right parenthesis end fraction
equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis A right parenthesis minus s i n left parenthesis 3 A right parenthesis plus s i n left parenthesis A right parenthesis over denominator cos left parenthesis 3 A right parenthesis minus cos left parenthesis 5 A right parenthesis plus cos left parenthesis 7 A right parenthesis plus cos left parenthesis 5 A right parenthesis end fraction
equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis 3 A right parenthesis over denominator cos left parenthesis 3 A right parenthesis plus cos left parenthesis 7 A right parenthesis end fraction
equals fraction numerator 2 s i n open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses over denominator 2 cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses end fraction
equals fraction numerator s i n 2 A over denominator cos 2 A end fraction
equals tan 2 A
equals R H S

Question 8(ix)

Solution 8(ix)

Question 8(x)

Solution 8(x)

Question 8(xi)

Solution 8(xi)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

Solution 13(i)

Question 13(ii)

Solution 13(ii)

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 19

If text   end text straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text   end text prove text   end text that text   end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana

Solution 19

table attributes columnalign left end attributes row cell Given text   that   end text straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text   end text end cell row cell text We   need   to   end text prove text   end text that text    end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana end cell row cell straight m text   end text sinθ equals text   end text straight n text   end text sin left parenthesis straight theta plus 2 straight a right parenthesis end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis over denominator text   end text sinθ end fraction equals straight m over straight n end cell row cell text Using   Componendo-Dividendo ,  we   have , end text end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis plus sinθ over denominator sin left parenthesis straight theta plus 2 straight a right parenthesis minus sinθ end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction.... left parenthesis 1 right parenthesis end cell row cell We text   know   that , end text end cell row cell text sinC + sinD = 2 sin end text fraction numerator text C + D end text over denominator text 2 end text end fraction text cos end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row and row cell text sinC end text minus text sinD  =  2 cos end text fraction numerator text C + D end text over denominator text 2 end text end fraction text sin end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row cell text Applying   the   above   formulae   in   equation  ( 1 ),  we   have , end text end cell row cell fraction numerator text 2 sin end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text cos end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction over denominator text 2 cos end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text sin end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text 2 sin end text left parenthesis straight theta plus straight a right parenthesis cosa over denominator text 2 cos end text left parenthesis straight theta plus straight a right parenthesis sina end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text tan end text left parenthesis straight theta plus straight a right parenthesis over denominator tana end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow text tan end text left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction cross times tana end cell row cell text Hence   proved. end text end cell end table

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RD SHARMA SOLUTION CHAPTER- 7 Trigonometric Ratios of Compound Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Ratios of Compound Angles Exercise Ex. 7.1

Question 34

I f space tan space left parenthesis A plus B right parenthesis equals p comma space tan space left parenthesis A minus B right parenthesis equals q comma space t h e n space s h o w space t h a t space tan space 2 A equals fraction numerator p plus q over denominator 1 minus p q end fraction

Solution 34

R H S comma
fraction numerator p plus q over denominator 1 minus p q end fraction
equals fraction numerator tan left parenthesis A plus B right parenthesis plus tan thin space left parenthesis A minus B right parenthesis over denominator 1 minus tan left parenthesis A plus B right parenthesis. tan left parenthesis A minus B right parenthesis end fraction
equals fraction numerator begin display style fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction end style plus begin display style fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end style over denominator 1 minus fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction. fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end fraction
equals fraction numerator begin display style fraction numerator left parenthesis tan thin space A plus tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis plus left parenthesis tan space A minus tan space B right parenthesis left parenthesis 1 minus tan space A. space tan thin space thin space B right parenthesis over denominator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style over denominator begin display style fraction numerator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis minus left parenthesis tan space A plus tan space B right parenthesis. left parenthesis tan space A minus tan space B right parenthesis over denominator left parenthesis 1 minus tan space A. space tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style end fraction
equals fraction numerator tan space A plus tan space B plus tan squared A. tan space B plus tan space A. tan squared B plus tan space A minus tan thin space B minus tan thin space squared A. tan space B plus tan space A. tan squared B over denominator 1 minus tan squared space A. tan squared space B minus tan squared space A plus tan squared space B end fraction
equals fraction numerator 2 space tan space A plus 2 space tan space A. tan squared B over denominator left parenthesis 1 minus tan squared A right parenthesis left parenthesis 1 plus tan squared B right parenthesis end fraction equals fraction numerator 2 space tan space A left parenthesis 1 plus tan squared space B right parenthesis over denominator left parenthesis 1 minus tan squared space A right parenthesis left parenthesis 1 plus tan squared space B right parenthesis end fraction equals fraction numerator 2 space tan space A over denominator 1 minus tan squared space A end fraction equals tan space 2 A equals L H S
H e n c e space P r o v e d

Question 19

I f space fraction numerator sin space left parenthesis x plus y right parenthesis over denominator sin space left parenthesis x minus y right parenthesis end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma space s h o w space t h a t space fraction numerator tan space x over denominator tan space y end fraction equals a over b

Solution 19

fraction numerator sin space x. cos space y plus sin space y. cos space x over denominator sin space x. cos space y minus sin space y. cos space x end fraction equals fraction numerator a plus b over denominator a minus b end fraction
rightwards double arrow fraction numerator sin space x. cos space y plus sin space y. cos space x plus sin space x. cos space y minus sin space y. cos space x over denominator sin space x. cos space y plus sin space y. cos space x minus sin space x. cos space y plus sin space y. cos space x end fraction equals fraction numerator a plus b plus a minus b over denominator a plus b minus a plus b end fraction left square bracket U sin g space C o m p o n e n d o space a n d space D i v i d e n d o right square bracket
rightwards double arrow fraction numerator 2 sin space x. cos space y over denominator 2 sin space y. cos space x end fraction equals fraction numerator 2 a over denominator 2 b end fraction
rightwards double arrow fraction numerator tan space x over denominator tan space y end fraction equals a over b
H e n c e space P r o v e d

Question 32

I f space a n g l e space theta space i s space d i v i d e d space i n t o space t w o space p a r t s space s u c h space t h a t space t h e space tan g e n t s space o f space o n e space p a r t space i s space lambda space p a r t
t i m e s space t h e space tan g e n t space o f space t h e space o t h e r comma space a n d space Ï• space i s space t h e i r space d i f f e r e n c e space t h e n space s h o w space t h a t space space
sin space theta equals fraction numerator lambda plus 1 over denominator lambda minus 1 end fraction sin space Ï•.

Solution 32

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 33

I f space tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction comma space t h e n space s h o w space t h a t space sin space alpha plus cos space alpha equals square root of 2 cos space theta

Solution 33

tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction
rightwards double arrow tan space theta equals space fraction numerator tan space alpha minus 1 over denominator tan space alpha plus 1 end fraction left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r space b y space cos space alpha right square bracket
rightwards double arrow tan space theta equals fraction numerator tan space alpha minus tan space begin display style pi over 4 end style over denominator 1 plus tan space begin display style pi over 4 end style. tan space alpha end fraction
rightwards double arrow tan space theta equals tan space open parentheses alpha minus pi over 4 close parentheses
rightwards double arrow theta equals alpha minus pi over 4 space space space left square bracket R e m o v i n g space tan space f r o m space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos open parentheses alpha minus pi over 4 close parentheses space left square bracket T a k i n g space cos space o n space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos space alpha. cos space pi over 4 plus sin space alpha. sin space pi over 4
rightwards double arrow cos space theta equals cos space alpha. fraction numerator 1 over denominator square root of 2 end fraction plus sin space alpha. fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow cos space theta equals fraction numerator cos space alpha plus sin space alpha over denominator square root of 2 end fraction
rightwards double arrow square root of 2 space cos space theta equals sin space alpha plus cos space alpha
H e n c e space P r o v e d

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(a)

Solution 2(a)

Question 2(b)

Solution 2(b)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 12(i)

Prove that:

Sin (60o – q) cos (30o + q) + cos (60o – q) sin (30o + q) = 1Solution 12(i)

Question 12(iii)

Prove that:

Solution 12(iii)

Question 12(ii)

Prove that:

Solution 12(ii)

Question 13

Solution 13

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 17(i)

Solution 17(i)

Question 17(ii)

Solution 17(ii)

Question 17(iii)

Solution 17(iii)

Question 17(iv)

Solution 17(iv)

Question 18

Solution 18

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 29(i)

Solution 29(i)

Question 29(ii)

Solution 29(ii)

Question 29(iii)

Solution 29(iii)

Question 30

Solution 30

Question 31

Solution 31

Chapter 17 Trigonometric Ratios of Compound Angles Exercise Ex. 7.2

Question 4

Solution 4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

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RD SHARMA SOLUTION CHAPTER-6 Graphs of Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 2(i)

Sketch the graph of the following pairs of functions on the same axes:

  1. y = sin x, y = sin begin mathsize 11px style open parentheses straight x plus straight pi over 4 close parentheses end style

Solution 2(i)

Question 2(ii)

Sketch the graph of the following pairs of functions on the same axes:

ii.  y = sin x, y = sin 3xSolution 2(ii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Chapter 6 Graphs of Trigonometric Functions Exercise Ex. 6.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Sketch the graph of the following functions on the same scale.

y = tan x, y = tan2 xSolution 9

Question 10

Solution 10

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Chapter – 18 Wastewater Story | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -18 Wastewater Story

Q.1. Fill in the blanks:
(a) Cleaning of water is process of removing __________
(b) Wastewater released by houses is called __________.
(c) Dried __________ is used as manure.
(d) Drains get blocked by __________ and __________.
Ans.(a) pollutants (b) sewage (c) sludge (d) (d) chemicals, kitchen waste

Q.2. What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas.
Ans. Sewage is a liquid containing wastes which is disposed by households, industrial and agricultural activities in water. It is harmful to discharge untreated sewage into rivers or seas because it can pollute the whole sources of water. Sewage contains harmful substances and disease causing organisms. It is therefore dangerous to release untreated sewage in water.

Q.3. Why should oils and fats be not released in the drain? Explain.
Ans. Oils and fats should not be released in drains because they harden the soil in the pipes and block them. Fats get clogged in holes of the soil in the drain and block it. It does not allow the water to flow.

Q.4. Describe the steps involved in getting clarified water from wastewater.
Ans. Following steps are involved in the purification of water:


(i) Firstly all the physical impurities like stones, plastic bags, cans etc. are to be removed. It is done by passing the water through bar screens.
(ii) Then water is taken to grit and sand removal tank where impurities are removed by sedimentation.
(iii) Solid impurities and feaces etc. are collected from bottom of water. These impurities collected are called sludge.
(iv) Clarified water is cleaned of other impurities by aerator. All disease causing bacteria are removed by chlorination.

Q.5. What is sludge? Explain how it is treated.
Ans. Sludge is the collected solid waste from the wastewater during the treatment in water treatment plant. Sludge is decomposed in a separate tank by the anaerobic bacteria. Activated sludge is used as manure.

Q.6. Untreated human excreta is a health hazard. Explain.
Ans. Untreated human excreta can cause a lot of health related problems. It pollutes water, air and soil. The polluted water contain disease causing bacteria which can spread epidemics like cholera, meningitis etc.

Q.7. Name two chemicals used to disinfect water.
Ans. Chlorine and ozone

Q.8. Explain the Junction of bar screens in a wastewater treatment plant.
Ans. Bar screens clear the wastewater of all the physical impurities. Large size waste like napkins, plastics, cans etc. are removed from the wastewater through the bar screens.

Q.9. Explain the relationship between sanitation and disease.
Ans. Sanitation and disease are related to each other. Sanitation involves proper disposal of sewage and refuse from hou&e and public places. If sanitation is there, no disease will occur, but if sanitation is not there various types of disease will occur and spread. So sanitation should be kept to avoid disease.

Q.10. Outline your role as an active citizen in relation to sanitation.
Ans. As active citizen we should take care of our personal environmental sanitation. We should make people aware of the benefits of sanitation. We should help municipal corporations to cover all the open drains and remove disease causing substances thrown in open.

Q.11.Here is a crossword puzzle. Good luck!
NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story Q11
Across:
3. Liquid waste products
4. Solid waste extracted in sewage treatment
6. A word related to hygiene
8. Waste matter discharged from human body
Down:
1. Used water
2. A pipe carrying sewage
5. Micro-organisms which cause cholera
7. A chemical to disinfect water

Ans
NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story Q11.1

Q.12. Study the following statements about ozone:
(a) It is essential for breathing of living organisms.
(b) It is used to disinfect water.
(c) It absorbs ultraviolet rays.
(d) Its proportion in air is about 3%.
Which of these statements are correct?
(i) (a), (b) and (c) (ii) (b) and (c) (iii) (a) and (d) (iv) All four
Ans. (ii) (b) and (c)

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RD SHARMA SOLUTION CHAPTER- 5 Trigonometric Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 5 Trigonometric Functions Exercise Ex. 5.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Prove that cosθ(tanθ+2)(2 tanθ+1)=2 secθ+5 sinθSolution 16

Question 17

I f space x equals fraction numerator 2 space sin space theta over denominator 1 plus cos space plus sin space theta end fraction comma space t h e n space p r o v e space t h a t space fraction numerator 1 minus cos space plus sin space theta over denominator 1 plus sin space theta end fraction i s space a l s o space e q u a l space t o space x.

Solution 17

fraction numerator 2 space sin space theta over denominator 1 plus cos space theta plus sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus cos space theta plus sin space theta right parenthesis left parenthesis 1 minus cos space theta plus sin space theta right parenthesis end fraction equals x space space space left square bracket R a t i o n a l i z i n g space t h e space d e n o m i n a t o r right square bracket
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 minus cos space theta plus sin space theta right parenthesis over denominator left parenthesis 1 plus sin space theta right parenthesis squared minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta minus 2 space sin space theta cos space theta plus 2 space sin squared theta over denominator 1 plus sin squared space theta plus 2 sin space theta minus cos squared space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin squared space theta plus 2 space sin space theta end fraction equals x
rightwards double arrow fraction numerator 2 space sin space theta left parenthesis 1 plus cos space theta minus sin space theta right parenthesis over denominator 2 sin space theta left parenthesis 1 plus sin space theta right parenthesis end fraction equals x
rightwards double arrow fraction numerator 1 plus cos space theta space minus sin space theta over denominator 1 plus sin space theta end fraction equals x space left square bracket C a n c e l l i n g space t h e space 2 space sin space theta space i n space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r right square bracket
H e n c e space P r o v e d

Question 18

Solution 18

Question 19

I f space tan space theta equals a over b comma space t h e n space f i n d space t h e space v a l u e space o f space square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root.

Solution 19

square root of fraction numerator a plus b over denominator a minus b end fraction end root plus square root of fraction numerator a minus b over denominator a plus b end fraction end root
equals square root of fraction numerator begin display style a over b end style plus 1 over denominator begin display style a over b end style minus 1 end fraction end root plus square root of fraction numerator begin display style a over b end style minus 1 over denominator begin display style a over b end style plus 1 end fraction end root space left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space d e n o m i n a t o r space b y space b right square bracket
equals square root of fraction numerator tan space theta plus 1 over denominator tan space theta minus 1 end fraction end root plus square root of fraction numerator tan space theta minus 1 over denominator tan space theta plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space theta over denominator cos space theta end fraction end style plus 1 end fraction end root
equals square root of fraction numerator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style end fraction end root plus square root of fraction numerator begin display style fraction numerator sin space theta minus cos space theta over denominator cos space theta end fraction end style over denominator begin display style fraction numerator sin space theta plus cos space theta over denominator cos space theta end fraction end style end fraction end root
equals square root of fraction numerator sin space theta plus cos space theta over denominator sin space theta minus cos space theta end fraction end root plus square root of fraction numerator sin space theta minus cos space theta over denominator sin space theta plus cos space theta end fraction end root
equals fraction numerator sin space theta plus cos space theta plus sin space theta minus cos space theta over denominator square root of sin squared space theta minus cos squared space theta end root end fraction
equals fraction numerator 2 space sin space theta space over denominator square root of sin squared space theta space minus space cos squared space theta end root end fraction

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Syntax error from line 1 column 702 to line 1 column 707. Unexpected '</mi>'.

Solution 24

L H S space equals space a b plus a minus b plus 1
equals left parenthesis s e c space theta minus tan space theta right parenthesis left parenthesis cos e c space plus c o t space right parenthesis plus s e c space theta minus tan space theta minus cos e c space theta minus c o t space theta plus 1
equals open parentheses fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction close parentheses open parentheses fraction numerator 1 over denominator sin space theta end fraction plus fraction numerator cos space theta over denominator sin space theta end fraction close parentheses plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator cos space theta end fraction cross times fraction numerator cos space theta over denominator sin space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction cross times fraction numerator 1 over denominator sin space theta end fraction minus tan space theta cross times c o t space theta plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction plus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator 1 over denominator cos space theta end fraction minus 1 plus fraction numerator 1 over denominator cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator 1 over denominator sin space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction plus 1
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 over denominator sin space theta space cos space theta end fraction minus fraction numerator sin space theta over denominator cos space theta end fraction minus fraction numerator cos space theta over denominator sin space theta end fraction
equals fraction numerator 1 minus sin squared space theta minus cos squared space theta over denominator sin space theta. cos space theta end fraction
equals fraction numerator 1 minus left parenthesis cos squared space theta plus sin squared space theta right parenthesis over denominator sin space theta. space cos space theta end fraction
equals fraction numerator 1 minus 1 over denominator sin space theta. cos space theta end fraction equals 0 equals R H S. space H e n c e space P r o v e d

Question 25

Solution 25

Question 26(i)

Solution 26(i)

Question 26(ii)

Solution 26(ii)

Question 26(iii)

Solution 26(iii)

Chapter 5 Trigonometric Functions Exercise Ex. 5.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 5 Trigonometric Functions Exercise Ex. 5.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)

Solution 1(xiv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

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Chapter – 17 Forests Our Lifeline | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -17 Forests Our Lifeline

Q.1. Explain how animals dwelling in the forest help it grow and regenerate.
Ans. Animals help in growing and regenerating forests in many ways. Animals work as the cleaning agents in the forest. Microorganisms work on dead bodies of plants and animals and degenerate them. An’imals also help in pollination which helps in growing a number of plants. Herbivores helps the carnivores to grow as they serve as food for them. Thus flora and fauna mutually grow in the forest.

Q.2. Explain how forests prevent floods.
Ans. Forests can absorb a lot of water. The roots of the trees absorb the water and prevent it from flowing away. Roots of trees also help in percolation of water into the soil. This helps in preventing floods.

Q.3. What are decomposers? Name any two of them. What do they do in the forest?
Ans. Decomposers are the organisms which feed on the dead bodies of plants and animals. They clean the forests decaying dead bodies and replenishing the nutrients back to the forest soil, e.g.„ beetles and grubs.

Q.4. Explain the role of forest in maintaining the balance between oxygen and carbon dioxide in the atmosphere.
Ans. Plants release oxygen in the atmosphere during the process of photosynthesis. This oxygen is inhaled by the animals for respiration. During respiration, they release carbon-dioxide which is absorbed by plants. In this way the oxygen and carbon dioxide cycle goes on. Since forests contain a large number of plants, they help much in this cycle and maintain balance in nature.

Q.5. Explain why there is no waste in a forest.
Ans. There is no waste in the forest because decomposers convert all the dead bodies of the plants and animals into the humus which gets added to the soil. Thus, no waste remains.

Q.6. List five products we get from forests.
Ans. (i) We get medicines from forests.
(ii) We get gum from forests.
(iii) We get wood which is used for many purposes like making furniture, paper etc.
(iv) We get food for animals from forests.
(v) We get sealing wax from forests.

Q.7.Fill in the blank:
(a) The insects, butterflies, honeybees and birds help flowering plants in .
(b) A forest is a purifier of and .
(c) Herbs form the layer in the forest.
(d) The decaying leaves and animal droppings in a forest enrich the .
Ans. (a) pollination (b) water, air (c) lowest (d) soil as humus.

Q.8. Why should we worry about the conditions and issues related to forests far from us?
Ans. We should be worried about deforestation as it would lead to floods, increase in earth’s temperature, decreasing animals habitats and soil erosion. Damage to forests directly or indirectly affects human habitat and environment so it must be a matter of concern among us.

Q.9. Explain why there is a need of variety of animals and plants in a forest.
Ans. All plants and animals sustain the forest life and also C02 – 02 cycle goes on due to animals and plants. Animals convert the dead and decaying matters into humus and increase the fertility of soil, thus enhancing plant growth. All food chains and food webs need variety of plants and animals.

Q.10. In fig. 1 7.15 the artist has forgotten to put the labels and directions on the arrows. Mark the directions on the arrows and label the diagram using the following labels: clouds, rain, atmosphere, cabon dioxide, oxygen, plants, animals, soil, roots, water table.
NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline Q10

Q.11.Which of the following is not a forest product?
(i) Gum (ii) Plywood (iii) Sealing wax (iv) Kerosene
Ans.(iv) Kerosene

Q.12.Which of the following statements is not correct?
(i) Forests protect the soil from erosion.
(ii) Plants and animals in a forest are not dependent on one another.
(iii) Forests influence the climate and water cycle.
(iu) Soil helps forests to grow and regenerate.
Ans.(ii) Plants and animals in a forest are not dependent on one another.

Q.13.Micro-organisms act upon the dead plants to produce
(i) sand (ii) mushrooms (iii) humus (iu) wood
Ans. (i) Humus

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RD SHARMA SOLUTION CHAPTER- 4 Measurement of Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 4 Measurement of Angles Exercise Ex. 4.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

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Chapter – 16 Water A Precious Resource | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -16 Water A Precious Resource

Q.1. Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) The freshwater stored in the ground is much more than that present in the rivers and lakes of the world. (T/F)


(b) Water shortage is a problem faced only by people living in rural areas. (T/F


(c) Water from rivers is the only source for irrigation in the fields. (T/F)
(d) Rain is the ultimate source of water. (T/F)

Ans. (a) T (b) F (c) F (d) T

Q.2. Explain how groundwater is recharged.
Ans. The groundwater gets recharged through the process of infiltration. Infiltration means seeping of water from rivers and lakes into the empty spaces and cracks deep below the ground.

Q.3.There are ten tubewells in a lane of fifty houses. What could be the long-term impact on the water table?
Ans.The effect on the water table depends on the replenishment of underground water. Only five families will share a tubewell, the water used for daily domestic purposes will not effect the water table. But if there is shortage of rains, the water used by the family will not replenished and water table will fall down.

Q.4.You have been asked to maintain a garden. How will you minimise the use of water? „
Ans.To minimise the wastage of water we will use the drip irrigation which throws the water at the base of plants. We will check the leakages in the water pipes and arrange small pits for rainwater harvesting. The collected rainwater will be used later.

Q.5.Explain the factors responsible for the depletion of water table.
Ans.Various factors responsible for the depletion of water table are:
(i) Increased population: Demand of water has been increased by the increased population. As the number of humans increase, the consumption of water also increases.
(ii) Increasing industries: All industries need water. As the number of human population increase, the number of industries are also increased which definitely increases the consumption of water.


(iii) Lack of water conservation techniques: Main source of water on earth and for the underground water is rain. The water of the rain, if conserved can increase the ground water level. But this is not done due to lack of water conservative techniques.
(iv) Agricultural activities: India is a country which depends on agriculture. The land used for cultivation has increased. So, the consumption of water for agriculture has increased. Irregular rainfall has increased the consumption of groundwater. This has increased the depletion of groundwater.

Q.6.Fill in the blanks with the appropriate answers: .
(a) People obtain groundwater through and .
(b) Three forms of water are solid, and .
(c) The water bearing layer of the earth is .
(d) The process of water seepage into the ground is called .
Ans.(a) wells, hand pumps
(b) liquid, gas
(c) aquifer
(d) infiltration

Q.7.Which one of the following is not responsible for water shortage?
(i) Rapid growth of industries
(ii) Increasing population
(iii) Heavy rainfall
(iv) Mismanagement of water resources
Ans.(iii) heavy rainfall

Q.8.Choose the correct option. The total water
(i) in the lakes and rivers of the world remains constant.
(ii) under the ground remains constant.
(iii) in the seas’and oceans of the world remains constant.
(iv) of the world remains constant.
Ans.(iv) of the world remains constant.

Q.9.Make a sketch shoyving groundwater and water table. Label it.
Ans
NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource Q9

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Chapter – 15 Light | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter -15 Light

Q.1.Fill in the blanks:
(a) An image that cannot be obtained on a screen is called ___________ .
(b) Image formed by a convex is __________ always virtual and smaller in size.


(c) An image formed by a __________ mirror is always of the same size as that of the object.


(d) An image which can be obtained on a screen is called a __________ image.
(e) An image formed by a concave __________ cannot be obtained on a screen.
Ans.(a) virtual image (b) mirror (c) plane (d) real (e) lens

Q.2.Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) We can obtain an enlarged and erect image by a convex mirror. (T/F)
(b) A concave lens always form a virtual image. (T/F)
(c) We can obtain a real, enlarged and inverted image by a concave mirror. (T/F)
(d) A real image cannot be obtained on a screen. (T/F)
(e) A concave mirror always form a real image. (T/F)
Ans. a) F (b) T (c) T (d) F (e) F

Q.3. Match the items given in Column I with one or more items of Column II
NCERT Solutions Class 7 Science Chapter 15 Light Q3
Ans.
NCERT Solutions Class 7 Science Chapter 15 Light Q3.1

Q.4.State the characteristics of the image formed by a plane mirror.
Ans.(i) Plane mirror forms an erect image.
(ii) It forms a virtual image.
(iii) Size of the image is same as that of the object.
(iv)Image is formed at the same distance behind the mirror as the object stands in front of it.
(v) Image formed is a laterally inverted image i.e., right hand side of the object seems to be the left hand side and vice-versa.

Q.5.Find out the letters of English alphabet or any other language known to you in which the image formed in a plane mirror appears exactly like the letter itself. Discuss your findings.
Ans.Letters like A, H, I, M, O, T, U ,V, W etc. appear same when seen through a plane mirror

Q.6.What is a virtual image? Give one situation where a virtual image is formed.
Ans.The image which cannot be taken on a screen is called virtual image. When some object is placed very close to the concave mirror we don’t get any image on the white screen placed behind the mirror. Such image is called virtual image.

Q.7. State two differences between a convex and a concave lens.
NCERT Solutions Class 7 Science Chapter 15 Light Q7

Q.8. Give one use each of a concave and a convex mirror.
Ans. Use of concave mirror:
Concave mirror is used by dentists to examine the teeth.
Use of convex mirror:
Convex mirror is used as side view mirror in vehicles

Q.9. Which type of mirror can form a real image?
Ans. Concave mirror can form a real image.

Q.10. Which type of lens forms always a virtual image?
Ans. Concave lens always forms a virtual image.

Choose the correct option in Questions 11-13:

Q.11. A virtual image larger than the object can be produced by a
(i) concave lens (ii) concave mirror (iii) convex mirror (iv) plane mirror

Ans. (ii) concave mirror

Q.12. David is observing his image in a plane mirror. Die distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be
(i) 3 m (ii) 5 m (iii) 6 m (iv) 8 m

Ans. (iii) 6 m

Q.13. The rear view mirror of a car is a plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear mew mirror the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be
(i) 1 m/s (ii) 2 m/s (iii) 4 m/s {iv) 8 m/s

Ans. (ii) 4 m/s

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RD SHARMA SOLUTION CHAPTER-3 Functions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 3 Functions Exercise Ex. 3.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Express the function f: X ® R given by f (x) = x 3 + 1 as set of ordered pairs, where X = {-1, 0, 3, 9, 7}.Solution 18

Chapter 3 Functions Exercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

begin mathsize 12px style If space straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator 1 minus straight x end fraction comma space Show space that space straight f left square bracket straight f left curly bracket straight f left parenthesis straight x right parenthesis right curly bracket right square bracket equals straight x. end style

Solution 4

Question 5

Solution 5

Question 6

begin mathsize 11px style Find space colon left parenthesis straight i right parenthesis space straight f space left parenthesis 1 divided by 2 right parenthesis space space left parenthesis ii right parenthesis space straight f left parenthesis negative 2 right parenthesis space space left parenthesis iii right parenthesis space straight f left parenthesis 1 right parenthesis space left parenthesis iv right parenthesis space straight f left parenthesis square root of 3 right parenthesis space and space left parenthesis straight v right parenthesis space straight f left parenthesis square root of negative 3 end root right parenthesis. end style

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10

Solution 10

Question 11

Solution 11

Chapter 3 Functions Exercise Ex. 3.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(viii)

find the domain and range of f open parentheses x close parentheses space equals space square root of 9 minus x squared end rootSolution 3(viii)

Question 3(vii)

Find domain and range of f (x) = -|x|Solution 3(vii)

As |x|is defined for all real numbers, its domain is R and range is only negative numbers because, |x| is always positive real number for all real numbers and -|x| is always negative real numbers.

Chapter 3 Functions Exercise Ex. 3.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 4(vii)

Solution 4(vii)

Question 4(viii)

Solution 4(viii)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals 3 x minus 2
f : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals minus x plus 4
f : R minus open curly brackets 3 over 2 close curly brackets rightwards arrow R space d e f i n e d space b y space f over g left parenthesis x right parenthesis equals fraction numerator x plus 1 over denominator 2 x minus 3 end fraction

Question 9

Solution 9

f plus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals square root of x plus x ; space
f minus g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals square root of x minus x ;
f g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals x to the power of 3 divided by 2 end exponent ;
f over g : left square bracket 0 comma infinity right parenthesis rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of x end fraction ;

Question 10

Solution 10

left parenthesis f plus g right parenthesis : R rightwards arrow left square bracket 0 comma infinity right parenthesis space d e f i n e d space b y space left parenthesis f plus g right parenthesis left parenthesis x right parenthesis equals x squared plus 2 x plus 1 equals left parenthesis x plus 1 right parenthesis squared
left parenthesis f minus g right parenthesis : R rightwards arrow R space d e f i n e d space b y space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals x squared minus 2 x minus 1
open parentheses f g close parentheses : R rightwards arrow R space d e f i n e d space b y space left parenthesis f g right parenthesis left parenthesis x right parenthesis equals 2 x cubed plus x squared
open parentheses f over g close parentheses : R rightwards arrow R space d e f i n e d space b y space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator x squared over denominator 2 x plus 1 end fraction
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