If tan ^{-1}left(frac{1-x}{1+x}right)=frac{1} | vedclass

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

Question

(C) Given the equation: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$ Using the identity $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) Given the equation: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$ Using the identity $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1}\left(\frac{a-b}{1+ab}ight)$,we can write $	an ^{-1}(1) - 	an ^{-1}(x) = \frac{1}{2} 	an ^{-1} x$ This simplifies to $\frac{\pi}{4} = 	an ^{-1} x + \frac{1}{2} 	an ^{-1} x$ $\frac{\pi}{4} = \frac{3}{2} 	an ^{-1} x$ $	an ^{-1} x = \frac{\pi}{4} 	imes \frac{2}{3} = \frac{\pi}{6}$ Therefore,$x = 	an \left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$,then $x$ is

Similar Questions

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

Considering only the principal values of an inverse function,the set $A = \{x \geq 0 \mid \tan^{-1} x + \tan^{-1} 6x = \frac{\pi}{4}\}$

The principal value of $\sin^{-1} \left( \sin \frac{5\pi}{3} \right)$ is

Find the value of $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$

If $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x - \tan ^{-1} x}{3 x^{3}}$ is equal to $L$,then the value of $(6L + 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module