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← IIT-JEE 2011 · Paper 1
MathematicsInteger+4Question 69 of 69

IIT-JEE 2011 · Paper 1, Mathematics question 69

Let f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ))f(\theta)=\sin\left(\tan^{-1}\left(\dfrac{\sin\theta}{\sqrt{\cos2\theta}}\right)\right), where −π4<θ<π4-\dfrac\pi4<\theta<\dfrac\pi4. Then the value of
dd(tan⁡θ)(f(θ))\frac{d}{d(\tan\theta)}\left(f(\theta)\right)
is

Answer and worked solution

Official answer: 1

Simplify. With u=sin⁡θcos⁡2θu=\frac{\sin\theta}{\sqrt{\cos2\theta}}: 1+u2=cos⁡2θ+sin⁡2θcos⁡2θ=cos⁡2θcos⁡2θ1+u^2=\dfrac{\cos2\theta+\sin^2\theta}{\cos2\theta}=\dfrac{\cos^2\theta}{\cos2\theta}, so
f(θ)=sin⁡(tan⁡−1u)=u1+u2=sin⁡θcos⁡θ=tan⁡θ.f(\theta)=\sin(\tan^{-1}u)=\frac u{\sqrt{1+u^2}}=\frac{\sin\theta}{\cos\theta}=\tan\theta.

Derivative with respect to tan⁡θ\tan\theta: 1.

Answer: 1

Topics:Inverse trigonometric functionsChain rule and implicit differentiation
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