If y = tan^{-1} left( frac{x}{1 + sqrt{1 - x^ | vedclass

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

Question

(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$.The first term is $y_1 = 	an^{-1} \left( \frac{\cos 	heta}{1 + \sin 	heta} ight) = 	an^{-

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$.The first term is $y_1 = 	an^{-1} \left( \frac{\cos 	heta}{1 + \sin 	heta} ight) = 	an^{-1} \left( \frac{\sin(\pi/2 - 	heta)}{1 + \cos(\pi/2 - 	heta)} ight) = 	an^{-1} \left( 	an \left( \frac{\pi/2 - 	heta}{2} ight) ight) = \frac{\pi}{4} - \frac{	heta}{2}$.The second term is $y_2 = \sin \left( 2 	an^{-1} \sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}} ight) = \sin \left( 2 	an^{-1} \left( 	an \frac{	heta}{2} ight) ight) = \sin 	heta = \sqrt{1 - x^2}$.Thus,$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x + \sqrt{1 - x^2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - x^2}} ight) + \frac{1}{2\sqrt{1 - x^2}} (-2x) = \frac{1}{2\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \frac{1 - 2x}{2\sqrt{1 - x^2}}$.

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

Similar Questions

If $\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3(x^3 - y^3)$,then $\frac{dy}{dx} = $

If $y = \tan^{-1}\left( \frac{a\cos x - b\sin x}{b\cos x + a\sin x} \right)$,then $\frac{dy}{dx} = $

If $y = \sin^{-1}(\sqrt{1 - x^2})$,then $dy/dx = $

If $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$,then $\frac{dy}{dx} = $

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module