Question

If $f(x)=\log _{\cot x} \tan x \cdot \log _{\tan x} \cot x^{-1}+$ $\tan ^{-1} \frac{x}{\sqrt{4-x^2}}, x \in(-2,2)$, then the value of $2 f^{\prime}(0)=$ ____

Answer 1

Let $a=\log _{\cot x} \tan x \log _{\tan x} \cot x^{-1}$
and $b=\tan ^{-1} \frac{x}{\sqrt{4-x^2}}$
$\Rightarrow f(x)=a+b$
$a=\log _{\cot x} \tan x \log _{\tan x} \cot x^{-1}$
$=\log _{\cot x} \tan x \frac{1}{\log _{\tan x} \cot x}$
$=\frac{\log (\tan x)^2}{\log (\cot x)}=\frac{\log (\tan x)^2}{\log \frac{1}{\tan x}}$
$=\frac{\log (\tan x)^2}{-\log (\tan x)}=1$
$b=\tan ^{-1} \frac{x}{\sqrt{4-x^2}}$
Let $\theta=\sin ^{-1} \frac{x}{2},-2< x< 2$
$\Rightarrow x=2 \sin \theta$
$\Rightarrow b=\tan ^{-1} \frac{2 \sin \theta}{\sqrt{4-4 \sin ^2 \theta}}$
$=\tan ^{-1} \frac{2 \sin \theta}{2 \cos \theta}$
$=\tan ^{-1}(\tan \theta),-\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$=\theta$
$=\sin ^{-1} \frac{x}{2}$
$\Rightarrow f(x)=1+\sin ^{-1} \frac{x}{2}$
$\Rightarrow f^{\prime}(x)=\frac{1}{\sqrt{1-\frac{x^2}{2}}} \cdot \frac{1}{2}$
$\Rightarrow f^{\prime}(x)=\frac{1}{\sqrt{4-x^2}}$
$\Rightarrow f^{\prime}(0)=\frac{1}{\sqrt{4-0}}=\frac{1}{2} \Rightarrow 2 f^{\prime}(0)=1$