RD SHARMA SOLUTION CHAPTER- 18 Binomial Theorem I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 18 Binomial Theorem Exercise Ex. 18.1

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 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

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 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 3

Solution 3

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 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

Show that 24n + 4 – 15n – 16, where n Î N is divisible by 225.Solution 12

Chapter 18 Binomial Theorem Exercise Ex. 18.2

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(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)

Question 9(vi)

Solution 9(vi)

Question 9(vii)

Solution 9(vii)

Question 9(viii)

Find the coefficient of x in the expansion of

 (1 – 3x + 7x2) (1 – x)16.Solution 9(viii)

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 13(iii)

Solution 13(iii)

Question 13(iv)

Solution 13(iv)

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 14(iii)

Solution 14(iii)

Question 14(iv)

Solution 14(iv)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Question 15(v)

Solution 15(v)

Question 15(vi)

Solution 15(vi)

Question 15(vii)

Solution 15(vii)

Question 15(viii)

Find the middle term (s) in expansion of:

Solution 15(viii)

Question 15(ix)

Find the middle term (s) in expansion of:

Solution 15(ix)

Question 15(x)

Find the middle term (s) in expansion of:

Solution 15(x)

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 16(vii)

Solution 16(vii)

Question 16(viii)

Solution 16(viii)

Question 16(ix)

Solution 16(ix)

Question 16(x)

Solution 16(x)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

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

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

If the seventh term from the beginning and end in the binomial expansion of   are equal, find n.Solution 39

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RD SHARMA SOLUTION CHAPTER-17 Combinations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 17 Combinations Exercise Ex. 17.1

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 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

If n + 2C8 : n – 2P4 = 57 : 16, find n.Solution 10

Question 11

Solution 11

Question 12

If nC2nC5 and nC6 are in A.P., then find nSolution 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

rightwards double arrow 42504

Question 20 (i)

Solution 20 (i)

Question 20(ii)

Solution 20(ii)

Question 20(iii)

Solution 20(iii)

Question 20(iv)

Solution 20(iv)

Chapter 17 Combinations Exercise Ex. 17.2

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

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

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Chapter 17 Combinations Exercise Ex. 17.3

Question 1

Solution 1

Question 3

Solution 3

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 2

Solution 2

Question 4

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.Solution 4

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RD SHARMA SOLUTION CHAPTER- 16 Permutations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 16 Permutations Exercise Ex. 16.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

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 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(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 12

Solution 12

Chapter 16 Permutations Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

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

How many three-digit numbers are there whit no digit repeated?Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19(i)

Solution 19(i)

Question 19(ii)

Solution 19(ii)

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?Solution 24Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.

One-digit odd number:

3 possible ways are there. These numbers are 3 or 5 or 7.

Two-digit odd number:

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3  2 = 6 such 2-digit numbers.

Three-digit odd number:

Ignore the presence of zero at ones place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3  3  2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1  3  2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 –  6 = 12.

Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

How many for digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4, if the digits can repeat?Solution 33

The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4  4  4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4  4  4  = 64

Now, consider four digit natural numbers whose digit at thousandths place is 4:

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.

Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4  4 + 4  4 + 4 + 1 = 37

Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.Question 34

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when on digit is repeated? How many of them are divisible by 10?Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

In how many ways can 5 different balls be distributed among three boxes?Solution 46

Question 47(i)

Solution 47(i)

Question 47(ii)

Solution 47(ii)

Question 47(iii)

Solution 47(iii)

Question 48

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.Solution 48

Each lamps has two possibilities either it can be switched on or off.

There are 10 lamps in the hall.

So the total numbers of possibilities are 210.

To illuminate the hall we require at least one lamp is to be switched on.

There is one possibility when all the lamps are switched off. If all the bulbs are switched off then hall will not be illuminated.

So the number of ways in which the hall can be illuminated is 210-1.

Chapter 16 Permutations Exercise Ex. 16.3

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(i)

Solution 1(i)

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

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

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.Solution 32

Chapter 16 Permutations Exercise Ex. 16.4

Question 1

Solution 1

Question 2

Solution 2

T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe space S T R A N G E apostrophe comma space i n c l u d i n g space 2 space v o w e l s space left parenthesis A comma E right parenthesis space a n d space 5 space c o n s o n a n t s space left parenthesis S comma T comma R comma N comma G right parenthesis.
left parenthesis i right parenthesis space C o n s i d e r i n g space 2 space v o w e l s space a s space o n e space l e t t e r comma space w e space h a v e space 6 space l e t t e r s space w h i c h space c a n space b e space a r r a n g e d space i n space to the power of 6 p subscript 6 equals 6 factorial space w a y s
A comma E space c a n space b e space p u t space t o g e t h e r space i n space 2 factorial space w a y s.

H e n c e comma space r e q u i r e d space space n u m b e r space o f space w o r d s space
equals space 6 factorial cross times 2 factorial
equals space 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1 cross times 2
equals 720 cross times 2 space
equals 1440.

left parenthesis i i right parenthesis space T h e space t o t a l space n u m b e r space o f space w o r d s space f o r m e d space b y space u sin g space a l l space t h e space l e t t e r s space o f space t h e space w o r d s space apostrophe S T R A N G E apostrophe
i s space to the power of 7 p subscript 7 equals 7 factorial
equals 7 cross times 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1
equals 5040.
S o comma space t h e space t o t a l space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space n e v e r space t o g e t h e r
equals T o t a l space n u m b e r space o f space w o r d s space minus space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space a l w a y s space space t o g e t h e r
equals 5040 minus 1440
equals 3600

left parenthesis i i i right parenthesis space T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space o u t space o f space t h e s e space l e t t e r s space apostrophe A apostrophe space a n d space apostrophe E apostrophe space a r e space t h e space v o w e l s.
T h e r e space a r e space 4 space o d d space p l a c e s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space T h e space t w o space v o w e l s space c a n space b e space a r r a n g e d space i n space to the power of 4 p subscript 2 space w a y s.
T h e space r e m a i n i n g space 5 space c o n s o n a n t s space c a n space b e space a r r a n g e d space a m o n g space t h e m s e l v e s space i n space to the power of 5 p subscript 5 space w a y s.
T h e space t o t a l space n u m b e r space o f space a r r a n g e m e n t s
equals to the power of 4 p subscript 2 cross times to the power of 5 p subscript 5
equals fraction numerator 4 factorial over denominator 2 factorial end fraction cross times 5 factorial
equals 1440

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

Chapter 16 Permutations Exercise Ex. 16.5

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 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

How many permulations of the letters of the world ‘MADHUBANI’ do not begain with M but end with I?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

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

Solution 28

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RD SHARMA SOLUTION CHAPTER- 15 Linear Inequations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 15 Linear Inequations Exercise Ex. 15.1

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

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

Solution 28

Chapter 15 Linear Inequations Exercise Ex. 15.2

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

Solve the system of equation fraction numerator 2 x minus 3 over denominator 4 end fraction minus 2 greater or equal than fraction numerator 4 x over denominator 3 end fraction minus 6 comma space 2 open parentheses 2 x plus 3 close parentheses less than 6 open parentheses x minus 2 close parentheses plus 10Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solve the system of equation 

Solution 21

Chapter 15 Linear Inequations Exercise Ex. 15.3

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

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 15 Linear Inequations Exercise Ex. 15.4

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

Chapter 15 Linear Inequations Exercise Ex. 15.5

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

Chapter 15 Linear Inequations Exercise Ex. 15.6

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(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

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 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Show that the following system of linear equations has no solution:

X + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.Solution 7

Question 8

Show that the solution set of the following system of linear inequalities is an unbounded region 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.Solution 8

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RD SHARMA SOLUTION CHAPTER- 14 Quadratic Equations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 14 Quadratic Equations Exercise Ex. 14.1

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

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

Chapter 14 Quadratic Equations Exercise Ex. 14.2

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)

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 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

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RD SHARMA SOLUTION CHAPTER- 12 Mathematical Induction I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 12 Mathematical Induction Exercise Ex. 12.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 12 Mathematical Induction Exercise Ex. 12.2

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

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 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

for all nNSolution 39

Question 42

Solution 42

Question 43

Solution 43

Question 29

Prove by the principle of mathematical induction

n3 – 7n + 3 is divisible by 3 for all n Î NSolution 29

Question 30

Prove by the principle of mathematical induction

1 + 2 + 22 +…. + 2n = 2n + 1 -1 for all n Î NSolution 30

Question 31

Prove by the principle of mathematical induction

Solution 31

Question 40

Prove that

cos a + cos (a + b) + cos (a + 2b) + …..+ cos (a + (n – 1)b)

Solution 40

Question 41

Solution 41

Question 44

Solution 44

Question 45

Prove that the number of subsets of a set containing n distinct elements is 2n for all n Î N.Solution 45

Question 46

A sequence a1, a2, a3, …….. is defined by letting a1 = 3 and ak = 7 ak-1 for all natural numbers k ³ 2. Show that an = 3.7n-1 for all n Î N.Solution 46

Question 47

Solution 47

Question 48

A sequence x0, x1, x2, x3, ……. is defined by letting x0 = 5 and xk = 4 + xk -1 for all natural number k. show that xn = 5 + 4n for all n Î N using mathematical induction.Solution 48

Question 49

Using principle of mathematical induction prove that

Solution 49

Question 50

The distributive law from algebra states that for all real numbers c, a1 and a2, we have c (a1 + a2) = ca1 + ca2

Use this law and mathematical induction to prove that, for all natural numbers, n ³ 2, if c (a1 + a2 + …. + an) = ca1 + ca2 + …+ can.Solution 50

Question 28

Solution 28

Question 35

Solution 35

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

Chapter 11 Trigonometric Equations Exercise Ex. 11.1

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)

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 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

F i n d space t h e space g e n e r a l space s o l u t i o n space o f space sin 2 theta plus cos theta equals 0

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

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(vii)

Solution 3(vii)

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(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

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 4(ix)

Solution 4(ix)

Question 6(v)

Solution 6(v)

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)

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RD SHARMA SOLUTION CHAPTER-10 Sine and Cosine Formulae and Their Applications I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 10 Sine and Cosine Formulae and Their Applications Exercise Ex. 10.1

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

In any triangle ABC, prove the following:

b sinB – c sinC = a sin (B – C)Solution 11

Question 12

In any triangle ABC, prove the following:

a2sin(B – C)= (b2 –c2)sinASolution 12

Question 13

Solution 13

Question 14

In any triangle ABC, prove the following:

a(sinB – sinC) + b (sinC – sinA) + c (sinA – sinB) = 0Solution 14

Question 15

Solution 15

Question 16

In any triangle ABC, prove the following:

a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A –cos2B) = 0Solution 16

Question 17

In any triangle ABC, prove the following:

b cosB + c cosC = a cos(B – C)Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

In any triangle ABC, prove the following:

a cosA + b cosB + c cosC= 2b sinA sinC= 2c sinA sinBSolution 22

Question 23

a(cos B cosC + cosA)= b(cos C cosA + cosB)= c(cos A cosB + cosC)Solution 23

Question 24

Solution 24

Question 25

In ΔABC prove that, if Ө be any angle, then b cosӨ = c cos(A – Ө) + a cos(C + Ө)Solution 25

Question 26

In a ΔABC, if sin2A + sin2B = sin2C, show that the triangle is right angled.Solution 26

Question 27

In any ΔABC, if a2, b2, c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.Solution 27

Question 28

The upper part of a broken over by the wind makes an angle of 300 with the ground and the distance from the root to the point where the top of the tree touches the ground is 15m. Using sine rule, find the height of the tree.Solution 28

Question 29

At the foot of a mountain the elevation of its summit is 450; after ascending 1000m towards the mountain up a slope of 300 inclination, the elevation is found to be 600. Find the height of the mountain.Solution 29

Question 30

A space p e r s o n space o b s e r v e s space t h e space a n g l e space o f space e l e v a t i o n space o f space t h e space p e a k space o f space a space h i l l space f r o m space a space
s t a t i o n space t o space b e space apostrophe alpha apostrophe. space H e space w a l k s space apostrophe c apostrophe space m e t r e s space a l o n g space a space s l o p e space i n c l i n e d space a t space t h e space a n g l e space apostrophe beta apostrophe
a n d space f i n d s space t h e space a n g l e space o f space e l e v a t i o n space o f space t h e space p e a k space o f space t h e space h i l l space t o space b e space apostrophe gamma apostrophe. space
S h o w space t h a t space t h e space h e i g h t space o f space t h e space p e a k space a b o v e space t h e space g r o u n d space i s space fraction numerator c cross times sin alpha cross times sin open parentheses gamma minus beta close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction.

Solution 30

C o n s i d e r space t h e space f o l l o w i n g space f i g u r e.

T h e space p e r s o n space i s space o b s e r v i n g space t h e space p e a k space P space f r o m space t h e space p o i n t space Q.
T h e space d i s tan c e space h e space t r a v e l l e d space i s space Q T equals c space m e t r e s space a n d space t h e space a n g l e space o f space i n c l i n a t i o n space o f
Q T space i s space beta.
H e space i s space o b s e r v i n g space t h e space p e a k space f r o m space t h e space p o i n t space a n d space t h e space a n g l e space o f space i n c l i n a t i o n space i s space gamma.
N o w space c o n s i d e r space t h e space t r i a n g l e space capital delta Q U T.
angle T Q U equals beta minus alpha
T h u s comma space sin open parentheses alpha minus beta close parentheses equals a over c
rightwards double arrow a equals c cross times sin open parentheses alpha minus beta close parentheses.... left parenthesis 1 right parenthesis
N o w space c o n s i d e r space t h e space t r i a n g l e space capital delta P Q R.
W e space k n o w space t h a t space angle Q P R equals 90 degree minus alpha
I n space t r i a n g l e space capital delta P T S comma space angle T P S equals 90 degree minus gamma
T h u s comma space angle T P U equals angle Q P R minus angle T P S
rightwards double arrow angle T P U equals open parentheses 90 degree minus alpha close parentheses minus open parentheses 90 degree minus gamma close parentheses
rightwards double arrow angle T P U equals gamma minus alpha
N o w space c o n s i d e r space t h e space capital delta T P U comma
T h u s comma space sin open parentheses gamma minus alpha close parentheses equals a over b
rightwards double arrow b equals fraction numerator a over denominator sin open parentheses gamma minus alpha close parentheses end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space a space f r o m space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e comma
b equals fraction numerator c cross times sin open parentheses alpha minus beta close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction... left parenthesis 2 right parenthesis
W e space n e e d space t o space f i n d space t h e space t o t a l space h e i g h t space o f space t h e space p e a k space P R.
H e r e comma space P R equals P S plus S R.... left parenthesis 3 right parenthesis
F r o m space t h e space t r i a n g l e space P S T comma space
sin gamma equals fraction numerator P S over denominator P T end fraction equals fraction numerator P S over denominator b end fraction
rightwards double arrow P S equals b sin gamma.... left parenthesis 4 right parenthesis
F r o m space t h e space t r i a n g l e space Q T W comma space
sin beta equals fraction numerator T W over denominator Q T end fraction equals fraction numerator T W over denominator c end fraction
rightwards double arrow T W equals S R equals c sin beta.... left parenthesis 5 right parenthesis
S u b s t i t u t i n g space t h e space v a l u e s space o f space P S space a n d space S R space f r o m space e q u a t i o n s space left parenthesis 4 right parenthesis space a n d space left parenthesis 5 right parenthesis
i n space e q u a t i o n space left parenthesis 3 right parenthesis comma space w e space h a v e
P R equals P S plus S R
rightwards double arrow P R equals b sin gamma plus c sin beta
rightwards double arrow P R equals fraction numerator c cross times sin open parentheses alpha minus beta close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction sin gamma plus c sin beta space space space space space space left square bracket f r o m space e q u a t i o n space left parenthesis 2 right parenthesis right square bracket
rightwards double arrow P R equals fraction numerator c cross times sin open parentheses alpha minus beta close parentheses cross times sin gamma plus c sin beta cross times sin open parentheses gamma minus alpha close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction
rightwards double arrow P R equals c open square brackets fraction numerator sin alpha cross times cos beta cross times sin gamma minus cos alpha cross times sin beta cross times sin gamma plus sin beta cross times sin gamma cross times cos alpha minus sin beta cross times sin alpha cross times cos gamma over denominator sin open parentheses gamma minus alpha close parentheses end fraction close square brackets
rightwards double arrow P R equals c open square brackets fraction numerator sin alpha cross times cos beta cross times sin gamma minus sin beta cross times sin alpha cross times cos gamma over denominator sin open parentheses gamma minus alpha close parentheses end fraction close square brackets
rightwards double arrow P R equals fraction numerator c sin alpha cross times open parentheses cos beta cross times sin gamma minus sin beta cross times cos gamma close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction
rightwards double arrow P R equals fraction numerator c sin alpha cross times sin open parentheses gamma minus beta close parentheses over denominator sin open parentheses gamma minus alpha close parentheses end fraction

Question 31

Solution 31

Chapter 10 Sine and Cosine Formulae and Their Applications Exercise Ex. 10.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 6

C (a cos B – b cos A) = a2 – b2Solution 6

Question 7

2(bc cos A + ca cos B +ab cosC)= a2 + b2 + c2Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 12

Solution 12

Question 13

Solution 13

Question 14

In a Δ ABC, prove that

sin3 A cos (B -C) + sin3B cos(C – A)+ sin3 C cos(A- B) = 3 sin A sin B sin CSolution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

I f space i n space a space capital delta A B C comma space cos squared A plus cos squared B plus cos squared C equals 1 comma space p r o v e space t h a t space t h e space t r i a n g l e space i s
r i g h t space a n g l e d.

Solution 17

C o n s i d e r space t h e space g i v e n space e q u a t i o n :
cos squared A plus cos squared B plus cos squared C equals 1
rightwards double arrow 1 minus sin squared A plus 1 minus sin squared B plus 1 minus sin squared C equals 1
rightwards double arrow 3 minus sin squared A plus 1 minus sin squared B plus 1 minus sin squared C equals 1

Question 18

Solution 18

Question 19

Solution 19

Question 5

b(c cos A – a cos C) = c2 –a2Solution 5

Question 11

In any DABC, prove the following:

a cos A + b cos B + c cosC = 2b sin A sin CSolution 11

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RD SHARMA SOLUTION CHAPTER- 9 Trigonometric Ratios of Multiple and Submultiple Angles I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.1

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

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Prove that

Solution 24

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 29

Solution 29

Question 30(i)

Solution 30(i)

Question 30(ii)

Solution 30(ii)

Question 30(iii)

Solution 30(iii)

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 36

Solution 36

Question 37

Solution 37

Question 38(i)

Solution 38(i)

Question 38(ii)

Solution 38(ii)

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 25

Solution 25

Question 31

Solution 31

Question 35

Solution 35

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.2

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

Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Exercise Ex. 9.3

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

Solution 9

Question 10

Solution 10

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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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