2 tan ^{-1}left(frac{1}{3}right)+tan ^{-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 $|x| < 1$. Applying this to the first term with $x = \frac{1}{3

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| Applying this to the first term with $x = \frac{1}{3}$: $2 	an ^{-1}\left(\frac{1}{3}ight) = 	an ^{-1}\left(\frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}ight) = 	an ^{-1}\left(\frac{2/3}{1-1/9}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{2}{3} 	imes \frac{9}{8}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$. Now,the expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{7}ight)$. Using the formula $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ where $xy $	an ^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}ight) = 	an ^{-1}\left(\frac{(21+4)/28}{1 - 3/28}ight) = 	an ^{-1}\left(\frac{25/28}{25/28}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

Similar Questions

If ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi ,$ then $\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} = $

If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$,then $5+x=$

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text { is equal to }\end{aligned}$

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

The value of $\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module