વિધેય $f(x) = \sqrt{|\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||}$ નો વિસ્તાર શોધો.

જો $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$ હોય,તો $\frac{dy}{dx} = $

જો $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$ હોય,તો $x = $

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) =$

જો $\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\cdots+\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right]=\tan ^{-1}[x]$ હોય,તો $x=$

$\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2(2)}-1}{\tan (2)}\right)-\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ ની કિંમત શોધો.