RD Sharma Class 10 Solutions Chapter 6 Exercise 6.1 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 6 Exercise 6.1: Students will use some basic trigonometric identities to prove useful trigonometric identities in this exercise. The RD Sharma Solutions Class 10 is the most important resource for students who want to clear their doubts and prepare for their exams with confidence. Students can download the RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities Exercise 6.1 PDF below for complete step-by-step solutions to exercise questions.

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RD Sharma Class 10 Solutions Chapter 6 Exercise 6.1

Access answers to RD Sharma Solutions Class 10 Maths Chapter 6 Exercise 6.1- Important Question with Answers

Prove the following trigonometric identities :
Question 1.
(1 – cos² A) cosec² A = 1
Solution:
(1 – cos² A) cosec² A = 1
L.H.S. = (1 – cos² A) cosec² A = sin² A cosec² A (∵ 1 – cos² A = sin² A)
= (sin A cosec A)² = (l)² = 1 = R.H.S. {sin A cosec A = 1 }

Question 2.
(1 + cot² A) sin² A = 1
Solution:
(1 + cot² A) sin² A = 1
L.H.S. = (1 + cot² A) (sin² A)
= cosec² A sin² A {1 + cot² A = cosec² A}
= [cosec A sin A]²
= (1)²= 1 = R.H.S. (∵ sin A cosec A = 1

Question 3.
tan² θ cos²  θ = 1- cos²  θ
Solution:

Question 4.

Solution:

Question 5.
(sec² θ – 1) (cosec² 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.
tan²  θ – sin²  θ = tan²  θ sin² θ
Solution:

Question 17.
(sec θ + cos θ ) (sec θ – cos θ ) = tan²  θ + sin² θ
Solution:

Question 18.
(cosec θ + sin θ) (cosec θ – sin θ) = cot² θ + cos² 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 + tan²  θ) (1 – sin θ) (1 + sin θ) = 1
Solution:

Question 22.
sin² A cot² A + cos² A tan² A = 1 (C.B.S.E. 1992C)
Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

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Question 28.

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Question 29.

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Question 30.

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Question 31.
sec⁶ θ= tan⁶  θ + 3 tan²  θ sec²  θ + 1
Solution:

Question 32.
cosec⁶  θ = cot⁶  θ+ 3cot²θ cosec²  θ + 1
Solution:

Question 33.

Solution:

Question 34.

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Question 35.

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Question 36.

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Question 37.

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Question 38.

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Question 39.

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Question 40.

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Question 41.

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Question 42.

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Question 43.

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Question 44.

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Question 45.

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Question 46.

Solution:

Question 47.


Solution:

Question 48.

Solution:

Question 49.
tan² A + cot² A = sec² A cosec² A – 2
Solution:

Question 50.

Solution:

Question 51.

Solution:

Question 52.

Solution:

Question 53.

Solution:

Question 54.
sin² A cos² B – cos² A sin² B = sin² A – sin² B.
Solution:
L.H.S. = sin² A cos² B – cos² A sin² B
= sin² A (1 – sin² B) – (1 – sin² A) sin² B
= sin² A – sin² A sin² B – sin² B + sin² A sin² B
= sin² A – sin² B
Hence, L.H.S. = R.H.S.

Question 55.

Solution:

Question 56.
cot² A cosec² B – cot² B cosec² A = cot² A – cot² B
Solution:

Question 57.
tan² A sec² B – sec² A tan² B = tan² A – tan² 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 x¹ – y² = a² – b¹. [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
x²-y² = {a sec θ + b tan θ)² – (a tan θ + b sec θ)²
= (a² sec² θ + b² tan²  θ + 2ab sec θ x tan θ) – (a² tan²  θ + b² sec²  θ + 2ab tan θ sec θ)
= a² sec² θ + b tan²  θ + lab tan θ sec θ – a² tan²  θ – b² sec²  θ – 2ab sec θ tan θ
= a² (sec² θ – tan²  θ) + b² (tan²  θ – sec²  θ)
= a² (sec² θ – tan²  θ) – b² (sec²  θ – tan²  θ)
= a² x 1-b² x 1 =a²-b² = 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.
sec⁴ A (1 – sin⁴ A) – 2tan² A = 1
Solution:

Question 74.

Solution:

Question 75.

Solution:

Question 76.

Solution:

Question 77.
If cosec θ – sin θ = a³, sec θ – cos θ = b³, prove that a²b² (a² + b²) = 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 a² + b² = m² + n²
Solution:

Question 81.
If cos A + cos² A = 1, prove that sin² A + sin⁴ A = 1
Solution:
cos A + cos² A = 1
⇒ cos A = 1 – cos² A
⇒cos A = sin² A
Now, sin² A + sin⁴ A = sin² A + (sin² A)²
= cos A + cos² A = 1 = R.H.S.

Question 82.
If cos θ + cos²  θ = 1, prove that
sin¹² θ + 3 sin¹⁰  θ + 3 sin⁸  θ + sin⁶  θ + 2 sin⁴  θ + 2 sin²  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:

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