The value of 2 tan ^{-1} frac{1}{2} + tan ^{- | vedclass

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

Question

(D) We know that $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.Applying this for $x = \frac{1}{2}$,we get $2 	an ^{-1} \frac{1}{2} =

Vedclass · Accepted Answer

$	an ^{-1} \left( \frac{41}{12} ight)$

Vedclass · Answer

(D) We know that $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.Applying this for $x = \frac{1}{2}$,we get $2 	an ^{-1} \frac{1}{2} = 	an ^{-1} \left( \frac{2 	imes (1/2)}{1 - (1/2)^2} ight) = 	an ^{-1} \left( \frac{1}{1 - 1/4} ight) = 	an ^{-1} \left( \frac{1}{3/4} ight) = 	an ^{-1} \left( \frac{4}{3} ight)$.Now,the expression becomes $	an ^{-1} \left( \frac{4}{3} ight) + 	an ^{-1} \left( \frac{3}{8} ight)$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$,we have:$	an ^{-1} \left( \frac{4/3 + 3/8}{1 - (4/3)(3/8)} ight) = 	an ^{-1} \left( \frac{(32+9)/24}{1 - 12/24} ight) = 	an ^{-1} \left( \frac{41/24}{12/24} ight) = 	an ^{-1} \left( \frac{41}{12} ight)$.

The value of $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{3}{8}$ is

Similar Questions

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$,where $-1 < x < 1$.

Show that $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$

If $\sin^{-1}(1 - x) - 2\sin^{-1}x = \pi/2$,then $x$ equals:

If $x = {\sin ^{ - 1}}(\sin 10)$ and $y = {\cos ^{ - 1}}(\cos 10)$,then $y - x$ is equal to

Find $\frac{dy}{dx}$ if $y = \sin^{-1}(2x\sqrt{1-x^2})$ where $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module