The value of tan ^{-1}left{frac{sqrt{1+x}-sqr | vedclass

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

Question

(B) Let $x = \cos 2	heta$. Then $	heta = \frac{1}{2} \cos^{-1} x$. Substitute $x = \cos 2	heta$ into the expression: $	an^{-1}\left(\frac{\sqrt{1+

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $x = \cos 2	heta$. Then $	heta = \frac{1}{2} \cos^{-1} x$. Substitute $x = \cos 2	heta$ into the expression: $	an^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta}+\sqrt{1-\cos 2	heta}}ight)$ $= 	an^{-1}\left(\frac{\sqrt{2\cos^2	heta}-\sqrt{2\sin^2	heta}}{\sqrt{2\cos^2	heta}+\sqrt{2\sin^2	heta}}ight)$ $= 	an^{-1}\left(\frac{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}ight)$ $= 	an^{-1}\left(\frac{1 - 	an	heta}{1 + 	an	heta}ight)$ $= 	an^{-1}(	an(\frac{\pi}{4} - 	heta))$ $= \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$. Now,adding the second term: $(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x) + \frac{1}{2} \cos^{-1} x = \frac{\pi}{4}$.

The value of $\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ is

Similar Questions

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots$ up to $100$ terms,then $\alpha$ is

The value of $x$ which satisfies $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$ is

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$ is

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

If for $|x|>1$, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$, then $f(-5)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module