RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities

RD Sharma Class 10 Solutions Chapter 6 Trigonometric Identities

RD Sharma Class 10 Solutions Trigonometric Identities Exercise 6.1

Prove the following trigonometric identities :
Question 1.
(1 – cos ² A) cosec ² A = 1
Solution:
(1 – cos ² A) cosec ² A = 1
L.H.S. = (1 – cos ² A) cosec ² A = sin ² A cosec ² A  (∵ 1 – cos ² A = sin ² A)
= (sin A cosec A) ² = (l) ² = 1 = R.H.S.  {sin A cosec A = 1 }

Question 2.
(1 + cot ² A) sin ² A = 1
Solution:
(1 + cot ² A) sin ² A = 1
L.H.S. = (1 + cot ² A) (sin ² A)
= cosec ² A sin ² A {1 + cot ² A = cosec ² A}
= [cosec A sin A] ²
= (1) ² = 1 = R.H.S. (∵ sin A cosec A = 1

Question 3.
tan ² θ cos ² θ = 1- cos ² θ
Solution:

Question 4.

Solution:

Question 5.
(sec ² θ – 1) (cosec ² θ – 1) = 1
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
tan ² θ – sin ² θ = tan ² θ sin ² θ
Solution:

Question 17.
(sec θ + cos θ ) (sec θ – cos θ ) = tan ² θ + sin ² θ
Solution:

Question 18.
(cosec θ + sin θ) (cosec θ – sin θ) = cot ² θ + cos ² θ
Solution:

Question 19.
sec A (1 – sin A) (sec A + tan A) = 1 (C.B.S.E. 1993)
Solution:

Question 20.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:

Question 21.
(1 + tan ² θ) (1 – sin θ) (1 + sin θ) = 1
Solution:

Question 22.
sin ² A cot ² A + cos ² A tan ² A = 1 (C.B.S.E. 1992C)
Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.
sec ⁶ θ= tan ⁶ θ + 3 tan ² θ sec ² θ + 1
Solution:

Question 32.
cosec ⁶ θ = cot ⁶ θ+ 3cot ² θ cosec ² θ + 1
Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.

Solution:

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Question 46.

Solution:

Question 47.


Solution:

Question 48.

Solution:

Question 49.
tan ² A + cot ² A = sec ² A cosec ² A – 2
Solution:

Question 50.

Solution:

Question 51.

Solution:

Question 52.

Solution:

Question 53.

Solution:

Question 54.
sin ² A cos ² B – cos ² A sin ² B = sin ² A – sin ² B.
Solution:
L.H.S. = sin ² A cos ² B – cos ² A sin ² B
= sin ² A (1 – sin ² B) – (1 – sin ² A) sin ² B
= sin ² A – sin ² A sin ² B – sin ² B + sin ² A sin ² B
= sin ² A – sin ² B
Hence, L.H.S. = R.H.S.

Question 55.

Solution:

Question 56.
cot ² A cosec ² B – cot ² B cosec ² A = cot ² A – cot ² B
Solution:

Question 57.
tan ² A sec ² B – sec ² A tan ² B = tan ² A – tan ² B
Solution:

Prove the following identities: (58-75)
Question 58.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x ¹ – y ² = a ² – b ¹ . [C.B.S.E. 2001, 20O2C]
Solution:
x – a sec θ + b tan θ
y = a tan θ + b sec θ
Squaring and subtracting, we get
x ² -y ² = {a sec θ + b tan θ) ² – (a tan θ + b sec θ) ²
= (a ² sec ² θ + b ² tan ² θ + 2ab sec θ x tan θ) – (a ² tan ² θ + b ² sec ² θ + 2ab tan θ sec θ)
= a ² sec ² θ + b tan ² θ + lab tan θ sec θ – a ² tan ² θ – b ² sec ² θ – 2ab sec θ tan θ
= a ² (sec ² θ – tan ² θ) + b ² (tan ² θ – sec ² θ)
= a ² (sec ² θ – tan ² θ) – b ² (sec ² θ – tan ² θ)
=  a ² x 1-b ² x 1 =a ² -b ² = R.H.S.

Question 59.
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ±3
Solutioon:

Question 60.
If cosec θ + cot θ = mand cosec θ – cot θ = n,prove that mn= 1
Solution:

Question 61.

Solution:

Question 62.

Solution:

Question 63.

Solution:

Question 64.

Solution:

Question 65.

Solution:

Question 66.
(sec A + tan A – 1) (sec A – tan A + 1) = 2 tan A
Solution:

Question 67.
(1 + cot A – cosec A) (1 + tan A + sec A) = 2
Solution:

Question 68.
(cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ-2)
Solution:

Question 69.
(sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
Solution:

Question 70.

Solution:

Question 71.

Solution:

Question 72.

Solution:

Question 73.
sec ⁴ A (1 – sin ⁴ A) – 2tan ² A = 1
Solution:

Question 74.

Solution:

Question 75.

Solution:

Question 76.

Solution:

Question 77.
If cosec θ – sin θ = a ³ , sec θ – cos θ = b ³ , prove that a ² b ² (a ² + b ² ) = 1
Solution:

Question 78.

Solution:

Question 79.

Solution:

Question 80.
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a ² + b ² = m ² + n ²
Solution:

Question 81.
If cos A + cos ² A = 1, prove that sin ² A + sin ⁴ A = 1
Solution:
cos A + cos ² A = 1
⇒ cos A = 1 – cos ² A
⇒cos A = sin ² A
Now, sin ² A + sin ⁴ A = sin ² A + (sin ² A) ²
= cos A + cos ² A = 1 = R.H.S.

Question 82.
If cos θ + cos ² θ = 1, prove that
sin ¹² θ + 3 sin ¹⁰ θ + 3 sin ⁸ θ + sin ⁶ θ + 2 sin ⁴ θ + 2 sin ² θ-2 = 1
Solution:

Question 83.
Given that :
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 – cos α) (l – cos β) (1 – cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Solution:

Question 84.

Solution:

Question 85.

Solution:

Question 86.
If sin θ + 2cos θ = 1 prove that 2sin θ – cos θ = 2. [NCERT Exemplar]
Solution:

RD Sharma Class 10th Solutions Chapter 6 Trigonometric Identities Ex 6.1

RD Sharma Class 10 Solutions

• Chapter 6 Trigonometric Identities 6.2