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JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

If 2 ^2((π)/(2) ^2 x)=1- (π 2 x), x ≠(2 n+1) (π)/(2), n ∈ I , then

If

2 \sin ^2\left(\frac{\pi}{2} \cos ^2 x\right)=1-\cos (\pi \sin 2 x), x \neq(2 n+1) \frac{\pi}{2}, n \in I

, then

Options

A $\cos 2 x$ is $\frac{3}{5}$
B $\cos 2 x$ is $\frac{1}{2}$
C $\tan x$ is $\frac{1}{2}$
D $\tan x$ is $\frac{1}{3}$

Correct answer

C. $\tan x$ is $\frac{1}{2}$

Step-by-step solution

\begin{aligned} & \text { We have } 2 \sin ^2\left(\frac{\pi}{2} \cos ^2 x\right)=2 \sin ^2\left(\frac{\pi}{2} \sin 2 x\right) \\ & \Rightarrow \cos ^2 x=\sin 2 x \Rightarrow \cos x(\cos x-2 \sin x)=0 \\ & \Rightarrow 1-2 \tan x=0 \text { as } \cos x \neq 0 \\ & \Rightarrow \tan x=\frac{1}{2} \text { and } \cos 2 x=\frac{3}{5}\end{aligned}

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