tan^{-1} frac{1}{2} + tan^{-1} frac{1}{3} = ? | vedclass

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

Question

(B) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - xy} ight)$ for $xy < 1$.Given $x = \frac{1}{2}$ and $y = \frac

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(B) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x + y}{1 - xy} ight)$ for $xy Given $x = \frac{1}{2}$ and $y = \frac{1}{3}$,we have $xy = \frac{1}{6} Substituting the values:$	an^{-1} \frac{1}{2} + 	an^{-1} \frac{1}{3} = 	an^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} 	imes \frac{1}{3}} ight)$$= 	an^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} ight)$$= 	an^{-1} \left( \frac{5/6}{5/6} ight)$$= 	an^{-1} (1)$$= \frac{\pi}{4}$.

$\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = ?$

Similar Questions

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is equal to

If $a=\sin ^{-1}(\sin (5))$ and $b=\cos ^{-1}(\cos (5))$,then $a^2+b^2$ is equal to

Find the derivative: $\frac{d}{dx} \cos^{-1} \left( \frac{x - x^{-1}}{x + x^{-1}} \right)$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module