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 $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ where $a, x \in(0,1)$,then the value of $x$ is

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

If $\frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right) = \tan^{-1} x$,then $x =$

If $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$,then $\sin x = $

$\cos^{-1}\left(\frac{15}{17}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module