2 tan ^{-1}left(frac{1}{3}right)+cos ^{-1}lef | vedclass

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

Question

(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| < 1$.First,calculate $2 	an ^{-1}\left(\frac{1}{3}ight)

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| First,calculate $2 	an ^{-1}\left(\frac{1}{3}ight) = 	an ^{-1}\left(\frac{2 	imes \frac{1}{3}}{1-(\frac{1}{3})^2}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.Now,convert $\cos ^{-1}\left(\frac{3}{5}ight)$ to $	an ^{-1}$. Let $	heta = \cos ^{-1}\left(\frac{3}{5}ight)$,then $\cos 	heta = \frac{3}{5}$. Since $\cos^2 	heta + \sin^2 	heta = 1$,$\sin 	heta = \sqrt{1 - (3/5)^2} = 4/5$. Thus,$	an 	heta = \frac{4/5}{3/5} = \frac{4}{3}$,so $\cos ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{4}{3}ight)$.The expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{4}{3}ight)$.Using $	an ^{-1}(x) + 	an ^{-1}(1/x) = \frac{\pi}{2}$ for $x > 0$,we get $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{3/4}ight) = \frac{\pi}{2}$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=$

Similar Questions

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

$\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x} = $

If $a_1, a_2, a_3, \dots, a_n$ are in arithmetic progression with common difference $d$, then $\tan [\tan^{-1} (\frac{d}{1 + a_1a_2}) + \tan^{-1} (\frac{d}{1 + a_2a_3}) + \dots + \tan^{-1} (\frac{d}{1 + a_{n-1}a_n})] = $

$\sin ^{-1}(\cos(\sin ^{-1} x)) + \cos ^{-1}(\sin(\cos^{-1} x)) = \text{ . . . . . . }$

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module