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# Let $f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right)} \right)$, where $\dfrac{{ - \pi }}{4} < \theta < \dfrac{\pi }{4}$. Then the value of $\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$ isA) 1B) 2C) 3D) 4  Hint: First of all we have to solve the function $f\left( \theta \right)$ by assuming ${\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right) = y$
Use the Pythagoras theorem to find out the necessary trigonometric ratios.

Given, $f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right)} \right)$
Let ${\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}} \right) = y$
$\Rightarrow \tan y = \dfrac{{\sin \theta }}{{\sqrt {\cos 2\theta } }}$ ….(1)
Now, $f\left( \theta \right) = \sin y$ ….(2)
Substitute $\cos 2\theta = 2{\cos ^2}\theta - 1$ in above equation (1),
$\Rightarrow \tan y = \dfrac{{\sin \theta }}{{\sqrt {2{{\cos }^2}\theta - 1} }} = \dfrac{{Perpendicular}}{{Base}}$
On applying Pythagoras Theorem,
${\left( {Hypo\tan eous} \right)^2} = {\left( {Perpendicular} \right)^2} + {\left( {Base} \right)^2}$
${\left( {Hypo\tan eous} \right)^2} = {\left( {\sin \theta } \right)^2} + {\left( {\sqrt {2{{\cos }^2}\theta - 1} } \right)^2} \\ {\left( {Hypo\tan eous} \right)^2} = {\sin ^2}\theta + 2{\cos ^2}\theta - 1 \\ {\left( {Hypo\tan eous} \right)^2} = \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) + {\cos ^2}\theta - 1 \\$
On putting the value of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ in above equation, we get,${\left( {Hypo\tan eous} \right)^2} = 1 + {\cos ^2}\theta - 1 \\$
${\left( {Hypo\tan eous} \right)^2} = {\cos ^2}\theta \\$
$Hypo\tan eous = \sqrt {{{\cos }^2}\theta } \\ Hypo\tan eous = \cos \theta \\$
We know that,
$\sin y = \dfrac{{Perpendicular}}{{Hypo\tan eous}}$\Rightarrow \sin y = \dfrac{{\sin \theta }}{{\cos \theta }} \Rightarrow \sin y = \dfrac{{\sin \theta }}{{\cos \theta }}$ \Rightarrow \sin y = \tan \theta$
$\Rightarrow \sin y = \tan \theta$
On putting the value of $\sin y$ in equation (2), we get
$f\left( \theta \right) = \tan \theta$
Now, $\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$= $\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {\tan \theta } \right)$
$\Rightarrow$$\dfrac{d}{{d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$= 1

Hence, option (A) is the correct answer.

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