Prove that, sin A - 2 sin 3 A 2 cos 2 A - cos A = tan A .

L.H.S = sin A - 2 sin 3 A 2 cos 2 A - cos A = sin A ( 1 - 2 sin 2 A ) cos A ( 2 cos 2 A - 1 ) = sin A ( 1 - 2 ( 1 - cos 2 A ) ) cos A ( 2 cos 2 A - 1 ) = sin A ( 1 - 2 + 2 cos 2 A ) ) cos A ( 2 cos 2 A - 1 ) = sin A ( 2 cos 2 A - 1 ) cos A ( 2 cos 2 A - 1 ) = sin A cos A = tan A = R.H.S Hence, proved.

If sin ⁡ θ + cos ⁡ θ = x , prove that sin 6 ⁡ θ + cos 6 ⁡ θ = 4 - 3 x 2 - 1 2 4 .

Given: sin ⁡ θ + cos ⁡ θ = x To prove: sin 6 ⁡ θ + cos 6 ⁡ θ = 4 - 3 x 2 - 1 2 4 On squaring the given equation, we get sin ⁡ θ + cos ⁡ θ 2 = x 2 ⇒ sin 2 ⁡ θ + 2 sin ⁡ θ cos ⁡ θ + cos 2 ⁡ θ = x 2 [ ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 ] ⇒ sin 2 ⁡ θ + cos 2 ⁡ θ + 2 sin ⁡ θ cos ⁡ θ = x 2 ⇒ 1 + 2 sin ⁡ θ cos ⁡ θ = x 2 ∵ sin 2 θ + cos 2 θ = 1 ⇒ 2 sin ⁡ θ cos ⁡ θ = x 2 - 1 ⇒ sin ⁡ θ cos ⁡ θ = x 2 - 1 2 . . . 1 Again, on squaring the above equation, we get sin ⁡ θ cos ⁡ θ 2 = x 2 - 1 2 4 ⇒ sin 2 ⁡ θ cos 2 ⁡ θ = x 2 - 1 2 4 . . . 2 Proof: By taking LHS, we get sin 6 ⁡ θ + cos 6 ⁡ θ = sin 2 ⁡ θ 3 + cos 2 ⁡ θ 3 = sin 2 ⁡ θ + cos 2 ⁡ θ 3 - 3 sin 2 ⁡ cos 2 ⁡ θ sin 2 ⁡ θ + cos 2 ⁡ θ ∵ a 3 + b 3 = a + b 3 - 3 a b = 1 3 - 3 x 2 - 1 2 4 1 ∵ sin 2 θ + cos 2 θ = 1 = 1 - 3 x 2 - 1 2 4 = 4 - 3 x 2 - 1 2 4 = RHS Hence proved.

If x = a sec ⁡ θ cos ⁡ ϕ , y = b sec ⁡ θ sin ⁡ ϕ and z = c tan ⁡ θ , Show that x 2 a 2 + y 2 b 2 - z 2 c 2 = 1 .

Given: x = a sec ⁡ θ cos ⁡ ϕ y = b sec ⁡ θ sin ⁡ ϕ To prove: x 2 a 2 + y 2 b 2 - z 2 c 2 = 1 . Solution: ⇒ x a = sec ⁡ θ cos ⁡ ϕ …(i) ⇒ y b = sec ⁡ θ sin ⁡ ϕ …(ii) ⇒ z c = tan ⁡ θ …(iii) On squaring the above equations and then subtracting the third from the sum of the first two, we get x a 2 + y b 2 - z c 2 = sec ⁡ θ cos ⁡ ϕ 2 + sec ⁡ θ sin ⁡ ϕ 2 - tan ⁡ θ 2 ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = sec 2 ⁡ θ cos 2 ⁡ ϕ + sec 2 ⁡ θ sin 2 ⁡ ϕ - tan 2 ⁡ θ ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = sec 2 ⁡ θ cos 2 ⁡ ϕ + sec 2 ⁡ θ sin 2 ⁡ ϕ - tan 2 ⁡ θ ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = sec 2 ⁡ θ cos 2 ⁡ ϕ + sin 2 ⁡ ϕ - tan 2 ⁡ θ ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = sec 2 ⁡ θ 1 - tan 2 ⁡ θ ∵ sin 2 ⁡ θ + cos 2 ⁡ θ = 1 ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = sec 2 ⁡ θ - tan 2 ⁡ θ ⇒ x 2 a 2 + y 2 b 2 - z 2 c 2 = 1 [ ∵ s e c 2 A - t a n 2 A = 1 ] = R H S Hence, proved.

E M B I B E

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1

R. D. Sharma Solutions for Chapter: Trigonometric Identities, Exercise 1: EXERCISE 11.1

Author:R. D. Sharma

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Trigonometric Identities, Exercise 1: EXERCISE 11.1

Attempt the free practice questions on Chapter 11: Trigonometric Identities, Exercise 1: EXERCISE 11.1 with hints and solutions to strengthen your understanding. MATHEMATICS CLASS X solutions are prepared by Experienced Embibe Experts.

Questions from R. D. Sharma Solutions for Chapter: Trigonometric Identities, Exercise 1: EXERCISE 11.1 with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1>Q 23

Prove that, $\frac{\sin A - 2 \sin^{3} A}{2 \cos^{2} A - \cos A} = \tan A$ .

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1>Q 80

If $a \cos θ + b \sin θ = m$ and $a \sin θ - b \cos θ = n,$ prove that $a^{2} + b^{2} = m^{2} + n^{2}$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1>Q 81

If $\cos A + \cos^{2} A = 1,$ prove that $\sin^{2} A + \sin^{4} A = 1$ .

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1>Q 82

If $\cos θ + \cos^{2} θ = 1,$ prove that $\sin^{12} θ + 3 \sin^{10} θ + 3 \sin^{8} θ + \sin^{6} θ + 2 \sin^{4} θ + 2 \sin^{2} θ - 2 = 1$ .

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 11 - Trigonometric Identities>EXERCISE 11.1>Q 83

Given that:
$(1 + \cos α) (1 + \cos β) (1 + \cos γ) = (1 - \cos α) (1 - \cos β) (1 - \cos γ)$
Show that one of the values of each member of this equality is $\sin α \sin β \sin γ$ .