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

(D) Given equation: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$Let $x = 	an 	heta$,then $	heta = 	an ^{-1} x$.Substituting $

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given equation: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$Let $x = 	an 	heta$,then $	heta = 	an ^{-1} x$.Substituting $x = 	an 	heta$ in the equation:$	an ^{-1}\left(\frac{1-	an 	heta}{1+	an 	heta}ight)=\frac{1}{2} 	heta$Using the formula $	an(\frac{\pi}{4} - 	heta) = \frac{1-	an 	heta}{1+	an 	heta}$:$	an ^{-1}\left(	an \left(\frac{\pi}{4}-	hetaight)ight)=\frac{1}{2} 	heta$$\frac{\pi}{4}-	heta=\frac{1}{2} 	heta$$\frac{\pi}{4}=\frac{3 	heta}{2}$$	heta=\frac{\pi}{6}$Since $x = 	an 	heta$,we have $x = 	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

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

Similar Questions

If $A = \tan^{-1}x$,then $\sin 2A = $

$2 \tanh^{-1} \frac{1}{2}$ is equal to

The simplest form of $\cot ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right), x>1$ is . . . . . . .

Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}, |x| < a$

If the range of $\operatorname{sech}^{-1} x + \operatorname{cosech}^{-1} x$ is $[a, b]$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module