RS Agarwal Solution | Class 10th | Chapter-13 | Trigonometric identities | Edugrown

Exercise Ex. 13A

Question 1

Prove the following identities:

Solution 1

(i)        

LHS = RHS

(ii)      

LHS = RHSQuestion 2

Prove the following identities:

(i)

(ii)

(iii)Solution 2

(i)

LHS = RHS

(ii)

LHS = RHS

(iii)

LHS = RHSQuestion 3

Prove the following identities:

(i)

(ii)Solution 3

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 4

Prove the following identities:

(i) 

_(ii) Solution 4

(i) 

(ii)

LHS = RHSQuestion 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 7

Prove the following identities:

(i)  

(ii)Solution 7

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 8

Prove the following identities:

Solution 8

(i)    LHS = 

(ii)             

Hence, LHS = RHSQuestion 9

Prove the following identity:

Solution 9

LHS = RHSQuestion 10

Prove the following identity:

Solution 10

Question 11

Prove the following identity:

Solution 11

Question 12

Prove the following identity:

Solution 12

RHS = LHSQuestion 13

Prove the following identity:

Solution 13

LHS = 

RHS = LHSQuestion 14

Prove the following identity:

Solution 14

LHS = RHSQuestion 15

Prove the following identity:

Solution 15

RHS = LHSQuestion 16

Prove the following identity:

Solution 16

Question 17

Prove the following identities:

Solution 17

(i)To prove 

     We know, 

      Therefore, LHS = RHS

(ii)

      Therefore, LHS = RHS

(iii)

Question 18

Prove the following identity:

(i) 

(ii) Solution 18

(i) 

LHS = RHS

(ii)

LHS = RHSQuestion 19

Prove the following identities:

(i) 

(ii)Solution 19

(i)

LHS = RHS

(ii) LHS =  

Question 20

Prove the following identities:

Solution 20

(i) 

   _{LHS =}

Hence, LHS = RHS

(ii)

LHS = RHSQuestion 21(i)

Solution 21(i)

Question 21(ii)

Solution 21(ii)

Question 21(iii)

Solution 21(iii)

Question 22

Prove the following identity:

Solution 22

LHS = RHSQuestion 23

Prove the following identity:

Solution 23

LHS = RHSQuestion 24

Prove the following identities:

(i)

(ii)Solution 24

(i)

LHS = RHS

(ii)

LHS = RHSQuestion 25

Prove the following identity:

Solution 25

LHS = RHSQuestion 26

Prove the following identities:

Solution 26

(i)

             Further,

                                LHS = RHS

(ii)

             LHS = 

Further,

Question 27

Prove the following identities:

(i) 

(ii)Solution 27

(i)

             On dividing the numerator and denominator of LHS by cos,We get

(ii)

            On dividing the numerator and denominator of LHS by cos,We get

                        LHS = RHS
Question 28

Prove the following identity:

Solution 28

LHS = RHSQuestion 29

Prove the following identity:

Solution 29

Question 30

Prove the following identity:

Solution 30

Question 31

Prove the following identity:

Solution 31

Question 32

Solution 32

Question 33

Prove the following identity:

Solution 33

Question 34

Prove the following identity:

Solution 34

Question 35

Prove the following identity:

Solution 35

Question 36(i)

Show that none of the following is an identity:

cos² θ + cos θ = 1Solution 36(i)

Question 36(ii)

sin² θ + sin θ = 2Solution 36(ii)

Question 36(iii)

tan² θ + sin θ = cos² θSolution 36(iii)

Question 37

Prove that:

(sin θ – 2sin³ θ) = (2cos³ θ – cos θ)tan θSolution 37

Exercise Ex. 13B

Question 1

If a cos + b sin = m and a sin – b cos = n, prove that

.Solution 1

Question 2

If x = a sec  + b tan and y = a tan  + b sec , prove that

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

If prove that Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If (cosec θ – sin θ) = a³ and (sec θ – cos θ) = b³, prove that a²b² (a² + b²) = 1.Solution 9

Question 10

If (2sin θ + 3cos θ) = 2, prove that (3sin θ – 2cos θ) = ± 3.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

If sec θ + tan θ = p, prove that

Solution 13(i)

Question 13(ii)

If sec θ + tan θ = p, prove that

Solution 13(ii)

Question 13(iii)

If sec θ + tan θ = p, prove that

Solution 13(iii)

Question 14

Solution 14

Question 15

Solution 15

Exercise Ex. 13C

Question 1

Write the value of (1 – sin² θ) sec² θ.Solution 1

Question 2

Write the value of (1 – cos²θ) cosec² θ.Solution 2

Question 3

Write the value of (1 + tan² θ) cos² θ.Solution 3

Question 4

Write the value of (1 + cot² θ) sin² θ.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Write the value of sin θ cos (90^ᵒ – θ) + cos θ sin (90^ᵒ – θ).Solution 7

Question 8

Write the value of cosec² (90^ᵒ – θ) – tan² θ.Solution 8

Question 9

Write the value of sec² θ (1 + sin θ)(1 – sin θ).Solution 9

Question 10

Write the value of cosec² θ(1 + cos θ)(1 – cos θ)

Note: Question modifiedSolution 10

Question 11

Write the value of sin² θ cos² θ (1 + tan² θ)(1 + cot² θ).Solution 11

Question 12

Write the value of (1 + tan² θ)(1 + sin θ)(1 – sin θ).Solution 12

Question 13

Write the value of 3cot² θ – 3cosec²θ.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

Write the value of tan 10ᵒ tan 20ᵒ tan 70ᵒ tan 80ᵒ.Solution 27

Question 28

Write the value of tan 1ᵒ tan 2ᵒ … tan 89ᵒ.Solution 28

Question 29

Write the value of cos 1ᵒ cos 2ᵒ…cos 180ᵒ.Solution 29

Question 30

Solution 30

Question 31

If sin θ = cos (θ – 45ᵒ),  where θ is a acute, find the value of θ.Solution 31

Question 32

Solution 32

Question 33

Find the value of sin 48ᵒ sec 42ᵒ + cos 48ᵒ cosec 42ᵒ .Solution 33

Question 34

If x = a sin θ and y = b cos θ, write the value of (b²x² + a²y²).Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

If sec θ + tan θ = x , find the value of sec θ.Solution 37

Question 38

Solution 38

Question 39

If sin θ = x, write the value of cot θ.Solution 39

Question 40

If sec θ = x, write the value of tan θ.Solution 40

Exercise MCQ

Question 1

Solution 1

Question 2

(a)0

(b) 1

(c) 2

(d) none of theseSolution 2

Question 3

tan 10° tan 15° tan 75° tan 80° = ?

Solution 3

Question 4

tan 5° tan 25° tan 30° tan 65° tan 85° = ?

Solution 4

Question 5

cos 1° cos 2° cos 3° …… cos 180° = ?

(a) -1

(b) 1

(c) 0

(d) Solution 5

Question 6

Solution 6

Question 7

sin 47° cos 43° + cos 47° sin 43° = ?

(a) sin 4° 

(b) cos 4° 

(c) 1

(d) 0Solution 7

Question 8

sec 70° sin 20° + cos 20° cosec 70° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 8

Question 9

If sin 3A = cos (A – 10^o) and 3A is acute then ∠A = ?

(a) 35° 

(b) 25° 

(c) 20° 

(d) 45° Solution 9

Question 10

If sec 4A = cosec (A – 10°) and 4A is acute then ∠A = ?

(a) 20° 

(b) 30° 

(c) 40° 

(d) 50° Solution 10

Question 11

If A and B are acute angles such that sin A = cos B then (A + B) =?

(a) 45° 

(b) 60° 

(c) 90° 

(d) 180° Solution 11

Question 12

If cos (𝛼 + 𝛽) = 0 then sin (𝛼 – 𝛽) = ?

(a) sin 𝛼 

(b) cos 𝛽 

(c) sin 2𝛼 

(d) cos 2𝛽 Solution 12

Question 13

sin (45° + θ) – cos (45° – θ) = ?

(a) 2 sin θ 

(b) 2 cos θ 

(c) 0

(d) 1Solution 13

Question 14

sec²10° – cot² 80° = ?

(a) 1

(b) 0

Solution 14

Question 15

cosec² 57° – tan² 33° = ?

(a) 0

(b) 1

(c) -1

(d) 2Solution 15

Question 16

Solution 16

Question 17

(a) 0

(b) 1

(c) 2

(d) 3Solution 17

Question 18

(a) 0

(b) 1

(c) -1

(d) none of theseSolution 18

Question 19

Solution 19

Question 20

(a) 30° 

(b) 45° 

(c) 60° 

(d) 90°Solution 20

Question 21

If 2cos 3θ = 1 then θ = ?

(a) 10° 

(b) 15° 

(c) 20° 

(d) 30° Solution 21

Question 22

(a) 15° 

(b) 30° 

(c) 45° 

(d) 60° Solution 22

Question 23

If tan x = 3cot x then x = ?

(a) 45° 

(b) 60° 

(c) 30° 

(d) 15° Solution 23

Question 24

If x tan 45° cos 60° = sin 60° cot 60° then x = ?

Solution 24

Question 25

If tan² 45° – cos² 30° = x sin 45° cos 45° then x = ?

Solution 25

Question 26

sec² 60° – 1 = ?

(a) 2

(b) 3

(c) 4

(d) 0Solution 26

Correct option: (b)

sec² 60° – 1 = (2)² – 1 = 4 – 1 = 3Question 27

(cos 0° + sin 30° + sin 45°)(sin 90° + cos 60° – cos 45°) =?

Solution 27

Question 28

sin²30° + 4cot² 45° – sec² 60° = ?

Solution 28

Question 29

3cos² 60° + 2cot² 30° – 5sin² 45° = ?

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

Solution 39

Question 40

Solution 40

Question 41

If (tan θ + cot θ) = 5 then (tan² θ + cot² θ) = ?

(a) 27

(b) 25

(c) 24

(d) 23Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

If sin A + sin² A = 1 then cos² A + cos⁴ A = ?

(a) 

(b) 1

(c) 2

(d) 3Solution 47

Question 48

If cos A + cos² A = 1 then sin² A + sin⁴ A = ?

(a) 1

(b) 2

(c) 4

(d) 3Solution 48

Question 49

(a) sec A + tan A

(b) sec A – tan A

(c) sec A tan A

(d) none of theseSolution 49

Question 50

(a) cosec A – cot A

(b) cosec A + cot A

(c) cosec A cot A

(d) none of theseSolution 50

Question 51

Solution 51

Question 52

(cosec θ – cot θ)² = ?

Solution 52

Question 53

(sec A + tan A)(1 – sin A) = ?

(a) sin A

(b) cos A

(c) sec A

(d) cosec ASolution 53

Exercise FA

Question 1

Solution 1

Question 2

Solution 2

Question 3

If cos A + cos² A = 1 then (sin² A + sin⁴ A) = ?

(a) 

(b) 2

(c) 1

(d) 4Solution 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

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.Solution 14

Question 15

If x = a sin θ + b cos θ and y = a cos θ – b sin θ, prove that x² + y² = a² + b².Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

If sec 5A = cosec (A – 36°) and 5A is an acute angle, show that A = 21° Solution 20