Question

If $y=tan(\frac{1}{2}co{s}^{-1}\frac{1-u^2}{1+u^2}+\frac{1}{2}si{n}^{-1}\frac{2u}{1+u^2})$ and $x=\frac{2u}{1-u^2}$, then $\frac{dy}{dx}$

Answer 1

Let us consider $\theta ϵ(-\frac{\pi }{2},\frac{\pi }{2})$ such that $u=\tan \theta$.

Now, $\frac{1-u^2}{1+u^2}=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$

and, $\frac{2u}{1{+u}^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$

Therefore, $y=\tan (\frac{1}{2}{\cos }^{-1}\frac{1-u^2}{1+u^2}+\frac{1}{2}{\sin }^{-1}\frac{2u}{1{+u}^2})$

$=\tan (\frac{1}{2}{\cos }^{-1}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+\frac{1}{2}{\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta })$

$=\tan (\frac{1}{2}{\cos }^{-1}\cos 2\theta +\frac{1}{2}{\sin }^{-1}\sin 2\theta )$ $($Using formulae of $\cos 2\theta$ and $\sin 2\theta$ $)$

$=\tan (\frac{1}{2}2\theta +\frac{1}{2}2\theta )$

$=\tan \theta$

$\Rightarrow y=u$

Therefore $\frac{dy}{dx}=1.$