Question

If $y=\tan (\frac{1}{2}{\cos }^{-1}\frac{1-u^2}{1+u^2}+\frac{1}{2}{\sin }^{-1}\frac{2u}{1+u^2})$ and $x=\frac{2u}{1-u^2}$ then $\frac{dy}{dx}$

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

Answer: (C)

$1$

$y=\tan (\frac{1}{2}{\cos }^{-1}\frac{1-u^2}{1+u^2}+\frac{1}{2}{\sin }^{-1}\frac{2u}{1+u^2})$ and $x=\frac{2u}{1-u^2}$
Put $u=\tan \theta$
$y=\tan (\frac{1}{2}{\cos }^{-1}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+\frac{1}{2}{\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta })$ and $x=\frac{2\tan \theta }{1-{\tan }^2\theta }$
$y=\tan \left(\frac{1}{2}{\cos }^{-1}\right(\cos 2\theta )+\frac{1}{2}{\sin }^{-1}\left(\sin 2\theta )\right)$ and $x=\tan 2\theta$
$y=\tan (\theta +\theta )$ and $x=\tan 2\theta$
$y=\tan 2\theta$ and $x=\tan 2\theta$
$\Rightarrow y=x$
$\frac{dy}{dx}=1$