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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Advanced 2011 · Paper 1 · Mathematics · Inverse Trigonometric Functions

Let f(θ)= [ ^ -1 ( θ 2 θ )] , where -(π)/(4) < θ < (π)/(4) . Then,...

Let f(θ)=sin⁡[tan⁡−1(sin⁡θcos⁡2θ)]f(\theta)=\sin \left[\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right], where −π4<θ<π4-\frac{\pi}{4} < \theta < \frac{\pi}{4}. Then, the value of dd(tan⁡θ)(f(θ))\frac{d}{d(\tan \theta)}(f(\theta)) isLet f(θ)= [ ^ -1 ( θ 2 θ )] , where -(π)/(4) < θ < (π)/(4) . Then, the value of (d)/(d( θ))(f(θ)) isLet f(θ)= [ ^ -1 ( θ 2 θ )] , where -(π)/(4) < θ < (π)/(4) . Then, the value of (d)/(d( θ))(f(θ)) is

Correct answer

0

Step-by-step solution

f(θ)=sin⁡(tan⁡−1sin⁡θcos⁡2θ),−π4<θ<π4f(\theta)=\sin \left(\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right),-\frac{\pi}{4} \lt \theta \lt \frac{\pi}{4} Let tan⁡−1sin⁡θcos⁡2θ=ϕ\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}=\phi \Rightarrow \quad \tan \phi=sin⁡θcos⁡2θ<imgsrc="/api/proxy−image?url=https\frac{\sin \theta}{\sqrt{\cos 2 \theta}}<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252FbbRBtOL8gHVb5FgMH9HblcPKawLYnY879QtTMljq-9M.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FbbRBtOL8gHVb5FgMH9HblcPKawLYnY879QtTMljq-9M.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2FbbRBtOL8gHVb5FgMH9HblcPKawLYnY879QtTMljq-9M.original.fullsize.png?alt=media"> \\ ∴sin⁡ϕ=sin⁡θsin⁡2θ+cos⁡2θ=sin⁡θ1−sin⁡2θ=sin⁡θcos⁡θ=tan⁡θ\begin{aligned} \therefore \quad \sin \phi& =\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}}\\ & =\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}=\frac{\sin \theta}{\cos \theta}=\tan \theta \end{aligned}∴f(θ)=sin⁡ϕ=tan⁡θ⇒df(θ)d(tan⁡θ)=1\begin{array}{ll} \therefore& f(\theta)=\sin \phi=\tan \theta\\ \Rightarrow & \frac{d f(\theta)}{d(\tan \theta)}=1 \end{array}
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