## 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