If cos x = tan y, cos y = tan z and cos z = tan x, then sin x = 2sin θ where θ is (where, x, y, z, θ are acute angles)

Correct answer: B. 18 ^ o ^ 2 x = ^ 2 y = ^ 2 y - 1 = ^ 2 z - 1 So, 1 + ^ 2 x = ^ 2 z = ^ 2 z ^ 2 z = ^ 2 z 1 - ^ 2 z = ^ 2 x 1 - ^ 2 x = ^ 2 x ^ 2 x - ^ 2 x ⇒ (2 - ^ 2 x) = ^ 2 x 1 - 2 ^ 2 x ⇒ (2 - ^ 2 x)(1 - 2 ^ 2 x) = ^ 2 x ⇒ 2 ^ 4 x - 6 ^ 2 x + 2 = 0 ⇒ ^ 2 x = 3 ± 9 - 4 2 = 3 ± 5 2 but ^ 2 x ∈ [0, 1] so rejecting ^ 2 x = 3 + 5 2 we have, ^ 2 x = 3 - 5 2 = ( 5 - 1 2 ) ^ 2 ⇒ sin x = 5 - 1 2 ⇒ 2sin θ = 5 - 1 2 ⇒ sin θ = 5 - 1 4 ⇒ θ = 18 ^ o

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If cos x = tan y, cos y = tan z and cos z = tan x, then sin x =...

cos x = tan y, cos y = tan z

and

cos z = tan x,

then

sin x = 2sin \theta

where

\theta

is (where,

x, y, z, \theta

are acute angles)

Options

A ${15}^{o}$
B ${18}^{o}$
C $22{\dfrac{1}{2}}^{o}$
D ${75}^{o}$

Correct answer

B. ${18}^{o}$

Step-by-step solution

\cos^{2}x = \tan^{2}y = \sec^{2}y - 1 = \cot^{2}z - 1

So,

1 + \cos^{2}x = \cot^{2}z = \dfrac{\cos^{2}z}{\sin^{2}z} = \dfrac{\cos^{2}z}{1 - \cos^{2}z}

= \dfrac{\tan^{2}x}{1 - \tan^{2}x} = \dfrac{\sin^{2}x}{\cos^{2}x - \sin^{2}x}

⇒ \left(2 - \sin^{2}x\right) = \dfrac{\sin^{2}x}{1 - 2\sin^{2}x}

⇒ \left(2 - \sin^{2}x\right)\left(1 - 2\sin^{2}x\right) = \sin^{2}x

⇒ 2\sin^{4}x - 6\sin^{2}x + 2 = 0

⇒ \sin^{2}x = \dfrac{3 \pm \sqrt{9 - 4}}{2} = \dfrac{3 \pm \sqrt{5}}{2}

but

\sin^{2}x \in \left[0, 1\right]

so rejecting

\sin^{2}x = \dfrac{3 + \sqrt{5}}{2}

we have,

\sin^{2}x = \dfrac{3 - \sqrt{5}}{2} = {\left(\dfrac{\sqrt{5} - 1}{2}\right)}^{2}

⇒ sin x = \dfrac{\sqrt{5} - 1}{2} ⇒ 2sin \theta = \dfrac{\sqrt{5} - 1}{2}

⇒ sin \theta = \dfrac{\sqrt{5} - 1}{4} ⇒ \theta = {18}^{o}

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