If y = sin left(2 tan ^{-1} sqrt{frac{1-x}{1+ | vedclass

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

Question

(B) Given $y = \sin \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight)$ and $x = \cos 2 	heta$. Substitute $x = \cos 2 	heta$ into the expression for

Vedclass · Accepted Answer

$-\cot 2 	heta$

Vedclass · Answer

(B) Given $y = \sin \left(2 	an ^{-1} \sqrt{\frac{1-x}{1+x}}ight)$ and $x = \cos 2 	heta$. Substitute $x = \cos 2 	heta$ into the expression for $y$: $\frac{1-x}{1+x} = \frac{1-\cos 2 	heta}{1+\cos 2 	heta} = \frac{2 \sin ^2 	heta}{2 \cos ^2 	heta} = 	an ^2 	heta$. Thus,$	an ^{-1} \sqrt{\frac{1-x}{1+x}} = 	an ^{-1}(	an 	heta) = 	heta$. Therefore,$y = \sin (2 	heta)$. Now,differentiate $y$ and $x$ with respect to $	heta$: $\frac{d y}{d 	heta} = \frac{d}{d 	heta} (\sin 2 	heta) = 2 \cos 2 	heta$. $\frac{d x}{d 	heta} = \frac{d}{d 	heta} (\cos 2 	heta) = -2 \sin 2 	heta$. Using the chain rule,$\frac{d y}{d x} = \frac{d y / d 	heta}{d x / d 	heta} = \frac{2 \cos 2 	heta}{-2 \sin 2 	heta} = -\cot 2 	heta$.

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

Similar Questions

If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}}, |t| \geq 1$,then $\frac{d y}{d x}=$

If $x = a(\cos \theta + \theta \sin \theta)$, $y = f(\theta)$, $f(2\pi) = 0$, $\frac{dy}{dx} = \frac{\tan \theta}{\theta}$, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right) = $

If $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$

If $u = \frac{\tan^{-1} x}{\tan^{-1} x + 1}$ and $v = \tan^{-1}(\tan^{-1} x)$,then $\frac{du}{dv} = \dots$

The differential coefficient of ${x^3}$ with respect to ${x^2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module