Question

diffrentiate 〖tan〗^(-1) (x/√(1-x^2 ))w.r.t.〖sec〗^(-1) (1/(〖2x〗^2-1)) *​

Answer 1

Let,

$\longrightarrow u=\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)$

Substitute,

$\longrightarrow x=\sin\theta$

Then,

$\longrightarrow u=\tan^{-1}\left(\dfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{\sin\theta}{\sqrt{\cos^2\theta}}\right)\quad\quad[\because\sin^2\theta+\cos^2\theta=1]$

$\longrightarrow u=\tan^{-1}\left(\dfrac{\sin\theta}{\cos\theta}\right)$

$\longrightarrow u=\tan^{-1}\left(\tan\theta\right)$

$\longrightarrow u=\theta$

$\longrightarrow\dfrac{du}{d\theta}=1\quad\quad\dots(1)$

Let,

$\longrightarrow v=\sec^{-1}\left(\dfrac{1}{2x^2-1}\right)$

$\longrightarrow v=\sec^{-1}\left(\dfrac{1}{2\sin^2\theta-1}\right)$

$\longrightarrow v=\sec^{-1}\left(-\dfrac{1}{1-2\sin^2\theta}\right)$

$\longrightarrow v=\sec^{-1}\left(\dfrac{1}{-\cos(2\theta)}\right)\quad\quad[\because\cos(2\theta)=1-2\sin^2\theta]$

$\longrightarrow v=\sec^{-1}\left(\dfrac{1}{\cos(\pi-2\theta)}\right)\quad\quad[\because\cos(\pi-\theta)=-\cos\theta]$

$\longrightarrow v=\sec^{-1}\left(\sec(\pi-2\theta)\right)$

$\longrightarrow v=\pi-2\theta$

$\longrightarrow\dfrac{dv}{d\theta}=-2\quad\quad\dots(2)$

Dividing (1) by (2) we get,

$\longrightarrow\underline{\underline{\dfrac{du}{dv}=-\dfrac{1}{2}}}$

This is the answer to the question.