begin{aligned} & text{If } y = tan^{-1} left{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that,$y = 	an^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - x}{1 + x}} ight\}$.Let $x = \

Vedclass · Accepted Answer

$\frac{1 - 2x}{2 \sqrt{1 - x^2}}$

Vedclass · Answer

(A) Given that,$y = 	an^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - x}{1 + x}} ight\}$.Let $x = \cos 2	heta$,then $	heta = \frac{1}{2} \cos^{-1} x$.Substituting $x = \cos 2	heta$ into the expression:$y = 	an^{-1} \left\{ \frac{\cos 2	heta}{1 + \sin 2	heta} ight\} + \sin \left\{ 2 	an^{-1} \sqrt{\frac{1 - \cos 2	heta}{1 + \cos 2	heta}} ight\}$$y = 	an^{-1} \left\{ \frac{\cos^2 	heta - \sin^2 	heta}{(\cos 	heta + \sin 	heta)^2} ight\} + \sin \left\{ 2 	an^{-1} (	an 	heta) ight\}$$y = 	an^{-1} \left\{ \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta} ight\} + \sin 2	heta$$y = 	an^{-1} \left\{ 	an \left( \frac{\pi}{4} - 	heta ight) ight\} + \sin 2	heta$$y = \frac{\pi}{4} - 	heta + \sin 2	heta$Substituting $	heta = \frac{1}{2} \cos^{-1} x$ and $\sin 2	heta = \sqrt{1 - x^2}$:$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x + \sqrt{1 - x^2}$Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - x^2}} ight) + \frac{1}{2\sqrt{1 - x^2}} (-2x)$$\frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \frac{1 - 2x}{2\sqrt{1 - x^2}}$.

$\begin{aligned} & \text{If } y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} \\ & + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\} \text{, then } \frac{dy}{dx} = \end{aligned}$

Similar Questions

If $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ where $|x| < 1$,then find the value of $\left(\frac{dy}{dx}\right)$ at $x = \frac{1}{2}$.

If $y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right)$,where $x^2 \le 1$. Then find $\frac{dy}{dx}$.

If $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} \right)$,then $\frac{dy}{dx}$ equals

If $y=\tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right)$,then $\frac{d y}{d x}=$

If $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$,then $\left(\frac{d y}{d x}\right)$ at $x=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module