If y = operatorname{cosec}^{-1}left[frac{sqrt | vedclass

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

Question

(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}ight) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Using the identity,we can rewrite the first term as $\sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Thus,$y = \sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}ight)$.Using the inverse trigonometric identity $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,we get $y = \frac{\pi}{2}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

If $y = \operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right] + \cos^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$,then $\frac{dy}{dx} = $

Similar Questions

If ${\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$,then $x =$

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$,where $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to

The value of $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module