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Question: Let \(f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\...

Let f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ))f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right)} \right), where −π4<θ<π4\dfrac{{ - \pi }}{4} < \theta < \dfrac{\pi }{4}. Then the value of dd(tan⁡θ)(f(θ))\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right) is
A) 1
B) 2
C) 3
D) 4

Explanation

Solution

First of all we have to solve the function f(θ)f\left( \theta \right) by assuming tan⁡−1(sin⁡θcos⁡2θ)=y{\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right) = y
Use the Pythagoras theorem to find out the necessary trigonometric ratios.

Complete step-by-step answer:
Given, f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ))f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right)} \right)
Let tan⁡−1(sin⁡θcos⁡2θ)=y{\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right) = y
⇒tan⁡y=sin⁡θcos⁡2θ\Rightarrow \tan y = \dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }} ….(1)
Now, f(θ)=sin⁡yf\left( \theta \right) = \sin y ….(2)
Substitute cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1 in above equation (1),
⇒tan⁡y=sin⁡θ2cos⁡2θ−1=PerpendicularBase\Rightarrow \tan y = \dfrac{{\sin \theta }}{{\sqrt {2{{\cos }^2}\theta - 1} }} = \dfrac{{Perpendicular}}{{Base}}
On applying Pythagoras Theorem,
(Hypotan⁡eous)2=(Perpendicular)2+(Base)2{\left( {Hypo\tan eous} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}
(Hypotan⁡eous)2=(sin⁡θ)2+(2cos⁡2θ−1)2 (Hypotan⁡eous)2=sin⁡2θ+2cos⁡2θ−1 (Hypotan⁡eous)2=(sin⁡2θ+cos⁡2θ)+cos⁡2θ−1  {\left( {Hypo\tan eous} \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\sqrt {2{{\cos }^2}\theta - 1} } \right)^2} \\\ {\left( {Hypo\tan eous} \right)^2} = {\sin ^2}\theta + 2{\cos ^2}\theta - 1 \\\ {\left( {Hypo\tan eous} \right)^2} = \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {\cos ^2}\theta - 1 \\\
On putting the value of sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 in above equation, we get,(Hypotan⁡eous)2=1+cos⁡2θ−1  {\left( {Hypo\tan eous} \right)^2} = 1 + {\cos ^2}\theta - 1 \\\

(Hypotan⁡eous)2=cos⁡2θ  {\left( {Hypo\tan eous} \right)^2} = {\cos ^2}\theta \\\ Hypotan⁡eous=cos⁡2θ Hypotan⁡eous=cos⁡θ  Hypo\tan eous = \sqrt {{{\cos }^2}\theta } \\\ Hypo\tan eous = \cos \theta \\\

We know that,
sin⁡y=PerpendicularHypotan⁡eous\sin y = \dfrac{{Perpendicular}}{{Hypo\tan eous}} ⇒sin⁡y=sin⁡θcos⁡θ \Rightarrow \sin y = \dfrac{{\sin \theta }}{{\cos \theta }}
⇒sin⁡y=sin⁡θcos⁡θ\Rightarrow \sin y = \dfrac{{\sin \theta }}{{\cos \theta }} ⇒sin⁡y=tan⁡θ\Rightarrow \sin y = \tan \theta
⇒sin⁡y=tan⁡θ\Rightarrow \sin y = \tan \theta
On putting the value of sin⁡y\sin y in equation (2), we get
f(θ)=tan⁡θf\left( \theta \right) = \tan \theta
Now, dd(tan⁡θ)(f(θ))\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)= dd(tan⁡θ)(tan⁡θ)\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {\tan \theta } \right)
⇒\Rightarrow dd(tan⁡θ)(f(θ))\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)= 1

Hence, option (A) is the correct answer.

Note: cos⁡2θ\cos 2\theta has three formulae i.e., cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta . Here, we use cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1, but we can also use the rest two i.e., cos⁡2θ−sin⁡2θ{\cos ^2}\theta - {\sin ^2}\theta or 2cos⁡2θ−12{\cos ^2}\theta - 1.