RS Aggarwal Solutions Chapter 8 Trigonometric Identities Exercise 8C Class 10 Maths

Chapter Name

RS Aggarwal Chapter 8 Trigonometric Identities

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 8A
  • Exercise 8B

Related Study

NCERT Solutions for Class 10 Maths

Exercise 8C Solutions

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

Solution

(1 – sin2 θ) sec2 θ

= cos2 θ × 1/cos θ

= 1


2. Write the value of (1 – cos2 θ) cosec2 θ.

Solution

(1 – cos2 θ) cosec2 θ

= sin2 θ × 1/sin2 θ

= 1


3. Write the value of (1 + tan2 θ) cos2θ.

Solution

(1 + tan2 θ) cos2 θ

= sec2θ × 1/sec2 θ

= 1


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

Solution

(1 + cot2 θ) sin2 θ

= cosec2 θ × 1/cosec2 θ

= 1


5. Write the value of (sin2 θ + 1+ tan2 θ).

Solution

(sin2 θ + (1 + tan2 θ)

= (sin2 θ + 1/sec2 θ)

= (sin2 θ + cos2 θ)

= 1


6. Write the value of (cot2 θ – 1/sin2 θ).

Solution

(cot2 θ – 1/sin2 θ)

= (cot2 θ – cosec2 θ)

= - 1


7. Write the value of sin θ cos(90° - θ) + cos θ sin (90° - θ).

Solution

sin θ cos(90° - θ) + cos θ sin(90° - θ)

= sin θ sin θ + cos θ cos θ

= sin2 θ + cos2 θ

= 1


8. Write the value of cosec2 (90° - θ) – tan2 - θ.

Solution

cosec2 (90° - θ) – tan2 θ

= sec2 θ – tan2 θ

= 1


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

Solution

sec2 θ(1 + sin θ)(1 – sin θ)

= sec2 θ(1 – sin2 θ)

= 1/cos2 θ × cos2 θ

= 1


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

Solution

cosec2 θ(1 + cos θ)(1 – cos θ)

= cosec2 θ(1 – cos2 θ)

= 1/sin2 θ × sin2 θ

= 1


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

Solution

sin2 θ cos2 θ(1 + tan2 θ)(1+ cot2 θ)

= sin2 θ cos2 θ sec2 θ cosec2 θ

= sin2 θ × cos2 θ × 1/cos2 θ × 1/sin2 θ

= 1


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

Solution

(1 + tan2 θ)(1 + sin θ)(1 – sin θ)

= sec2 θ(1 – sin2 θ)

= 1/cos2 θ × cos2 θ

= 1


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

Solution

3 cot2 θ – 3cosec2 θ

= 3(cot2 θ – cosec2 θ)

= 3(-1)

= - 3


14. Write the value of 3 cot2 θ – 3 cosec2 θ.

Solution

4 tan2 θ – 4/cos2 θ

= 4 tan2 θ – 4 sec2 θ

= 4(tan2 θ – sec2 θ)

= 4(-1)

= -4


15. Write the value of (tan2 θ – sec2 θ)/(cot2 θ – cosec2 θ).

Solution

(tan2 θ – sec2 θ)/(cot2 θ – cosec2 θ)

= -1/-1

= 1


16. If sin θ = 1/2, write the value of (3 cot2θ + 3).

Solution

As sin θ = 1/2

So, cosec θ = 1/sin θ = 2 ...(i)

Now.

3 cot2 θ + 3

= 3(cot2 θ + 1)

= 3 cosec2 θ

= 3(2)2 [Using (i)]

= 3(4)

= 12


17. If cos θ = 2/3, write the value of (4 + 4 tan2 θ).

Solution

4 + 4 tan2 θ

= 4(1 + tan2 θ)

= 4 sec2 θ

= 4/cos2 θ

= 4/(2/3)2

= 4/(4/9)

= (4 × 9)/4

= 9


18. If cos θ = 7/25, write the value of (tan θ + cot θ).

Solution

As sin2 θ = 1 – cos2 θ

= 1 – (7/25)2

= 1 – 49/625

= (625 – 49)/625

⇒ sin2 θ = 576/625

⇒ sin θ = 24/25

Now, tan θ + cot θ

= sin θ/cos θ + cos θ/sin θ

= (sin2 θ + cos2 θ)/(cos θ sin θ)

= 1/(7/25 × 24/25)

= 1/(168/625)

= 625/168


19. If cos θ = 2/3, write the value of (sec θ – 1)/(sec θ + 1).

Solution

(sec θ – 1)/(sec θ + 1)

= 1/5


20. If 5 tan θ = 4, write the value of (cos θ – sin θ)/(cos θ + sin θ).

Solution

We have,

5 tan θ = 4

⇒ tan θ = 4/5

Now,

(cos θ – sin θ)/(cos θ + sin θ)



21. If 3 cot θ = 4, write the value of (2 cos θ – sin θ)/(4 cos θ - sin θ).

Solution

We have:

3 cot θ = 4

⇒ cot θ = 4/5

Now,

(2 cos θ – sin θ)/(cos θ + sin θ)


= 11/13

22. If cot θ = 1/√3, write the value of (1 – cos2 θ)/(2 – sin2 θ).

Solution

We have:

Cot θ = 1/√3

⇒ cot θ = cot (π /3)

⇒ θ = π/3

Now, 


23. If tan θ = 1/√5, write the value of (cosec2 θ – sec2 θ)/(cosec2 θ – sec2 θ).

Solution

(cosec2 θ – sec2 θ)/(cosec2 θ – sec2 θ)


= 24/36
= 2/3

24. If cot A = 4/3 and (A + B) = 90°, what is the value of tan B?

Solution

We have:

cot A = 4/3

⇒ cot (90° - B) = 4/3 (As A + B = 90°)

∴ tan B = 4/3


25. If cos B = 3/5 and (A + B) = 90°, find the value of sin A.

Solution

We have,

cos B = 3/5

⇒ cos(90° - A) = 3/5 (As, A + B = 90°)

∴ sin A = 3/5


26. If √3 sin θ = cos θ and θ is an acute angle, find the value of θ.

Solution

We have,

√3 sin θ = cos θ

⇒ sin θ/cos θ = 1/√3

⇒ tan θ = 1/√3

⇒ tan θ = tan 30°

∴ θ = 30°


27. Write the value of tan 10° tan 20° tan 70° tan 80°.

Solution

tan 10° tan 20° tan 70° tan 80°

= cot (90° - 10°) cot (90° - 20°) tan 70° tan 80°

= cot 80° cot 70° tan 70° tan 80°

= 1/tan 80° × 1/tan 70° × tan 70° × tan 80°

= 1


28. Write the value of tan 1° tan 2° ...... tan 89°.

Solution

tan 1° tan 2° ...... tan 89°

= tan 1° tan 2° tan 3° ........ tan 45° ........ tan 87° tan 88° tan 89°

= tan 1° tan 2° tan 3° .......tan 45° .......cot(90° - 87°) cot(90° - 88°) cot(90° - 89°)

= tan 1° tan 2° tan 3° ..... tan 45° ....cot 3° cot 2° cot 1°

= tan 1° × tan 2° × tan 3° × ...... × 1 × ..... × 1/tan 3° × 1/tan 2° × 1/tan 1°

= 1


29. Write the value of cos 1° cos 2° .........cos 180°.

Solution

cos 1° cos 2° .......cos 180°

= cos 1° cos 2° .......cos 90° ......cos 180°

= cos 1° cos 2° ......0 ......cos 180°

= 0


30. If tan A = 5/12, find the value of (sin A + cos A) sec A.

Solution

(sin A + cos A sec A)

= (sin A + cos A) 1/cos A

= sin A/cos A + cos A/cos A

= tan A + 1

= 5/12 + 1/1

⇒ (5 + 12)/12

= 17/12


31. If sin θ cos (θ - 45°), where θ is acute, find the value of θ.

Solution

We have,

sin θ = cos (θ – 45°) θ

⇒ cos (90° - θ) = cos (θ - 45°)

Comparing both sides, we get

90° - θ = θ - 45°

⇒ θ + θ = 90° + 45°

⇒ 2θ = 135°

⇒ θ = (135/2)°

∴ θ = 67.5°


32. Find the value of sin 50°/cos 40° + cosec 40°/sec 50° - 4 cos 50° cosec 40°.

Solution

sin 50°/cos 40° + cosec 40°/sec 50° - 4 cos 50° cosec 40°

= cos(90° - 50°)/cos 40° + sec(90° - 40°)/sec 50° - 4 sin (90° - 50°) cosec 40°

= cos 40°/cos 40° + sec 50°/sec 50° - 4 sin 40° × 1/ sin 40°

= 1 + 1 – 4

= - 2


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

Solution

sin 48° sec 42° + cos 48° cosec 42°

= sin 48° cosec (90° - 42) + cos 48° sec (90° - 42°)

= sin 48° cosec 48° + cos 48° sec 48°.

= sin 48° × 1/ sin 48° + cos 48° × 1/cos 48°

= 1 + 1

= 2


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

Solution

(b2x2 + a2x2)

= b2 (a sin θ)2 + a2(b cos θ)2

= b2a2 sin2 θ + a2b2 cos2 θ

= a2b2 (sin2 θ + cos2 θ)

= a2b2(1)

= a2b2


35. If 5x = sec θ and 5/x = tan θ, find the value of 5(x2 – 1/x2).

Solution

5(x2 – 1/x2)

= 25/5(x2 – 1/x2)

= 1/5(25x2 – 25/x2)

= 1/5[(5x)2 – (5/x)2]

= 1/5[(sec θ)2 – (tan θ)2]

= 1/5(sec2 θ – tan2 θ)

= 1/5 (1)

= 1/5


36. If cosec θ = 2x and cot θ = 2/x, find the value of 2(x2 – 1/x2)

Solution

2(x2 – 1/x2)

= 4/2(x2 – 1/x2)

= 1/2(4x2 – 4/x2)

= 1/2[(2x2 – (2/x)2]

= 1/2[(cosec θ)2 – sec θ)2]

= 1/2(cosec2 θ – sec2 θ)

= 1/2(1)

= 1/2


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

Solution

We have:

sec θ + tan θ = x  ...(i)

⇒ (sec θ + tan θ)/1 × (sec θ - tan θ)/(sec θ - tan θ) = x

⇒ (sec2 θ - tan2 θ)/(sec θ - tan θ) = x

⇒ 1/(sec θ – tan θ) = x/1

 ⇒ sec θ – tan θ = 1/x ...(ii)

Adding (i) and (ii), we get

2 sec θ = x + 1/x

⇒ 2 sec θ = (x2 + 1)/x

∴ sec θ = (x2 + 1)/2x


38. Find the value of (cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)

Solution

(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)



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

Solution 

cot θ = cos θ/sin θ



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

Solution 

As tan2 θ = sec2 θ - 1
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