Explanation

(a)
$f(x) = (\log_{\cot x}{\tan{x)(\log_{\tan x}{\cot{x\left. \ \ \right)^{- 1}} + tan^{- 1}\frac{4x}{4 - x^{2}}}}}$
$= \frac{\log{\tan x}}{\log{\cot x}}.\frac{\log{\tan x}}{\log{\cot x}} + \tan^{- 1}\ \left( \frac{4x}{4 - x^{2}} \right)$
$= \frac{(\log{\tan{x\left. \ \ \right)^{2}}}}{\left( - \log{\tan x} \right)^{2}} + \tan^{- 1}\ \left( \frac{4x}{4 - x^{2}} \right)$
$= 1 + \tan^{- 1}\left( \frac{4x}{4 - x^{2}} \right)$
$\therefore f^{'}(x) = \frac{1}{1 + \left( \frac{4x}{4 - x^{2}} \right)^{2}}.\frac{\left( 4 - x^{2} \right)4 - 4x( - 2x)}{\left( 4 - x^{2} \right)^{2}}$
$= \frac{16 - 4x^{2} + 8x^{2}}{(4 - x)^{2} + 16x^{2}} = \frac{4(4 + x^{2})}{\left( 4 - x^{2} \right)^{2} + (4x)^{2}}$
Hence, $f(2) = \frac{4(4 + 4)}{0 + (8)^{2}} = \frac{32}{64} = \frac{1}{2}$