The solution of the equation tan ^{-1}(1+x)+t | vedclass

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

Question

(B) Given the equation: $	an ^{-1}(1+x)+	an ^{-1}(1-x)=\frac{\pi}{2}$ We know that $	an ^{-1}(A) + \cot ^{-1}(A) = \frac{\pi}{2}$,so $\frac{\pi}{2}

Vedclass · Accepted Answer

$x=0$

Vedclass · Answer

(B) Given the equation: $	an ^{-1}(1+x)+	an ^{-1}(1-x)=\frac{\pi}{2}$ We know that $	an ^{-1}(A) + \cot ^{-1}(A) = \frac{\pi}{2}$,so $\frac{\pi}{2} - 	an ^{-1}(1-x) = \cot ^{-1}(1-x)$. Substituting this into the equation: $	an ^{-1}(1+x) = \cot ^{-1}(1-x)$. Using the identity $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$,we get: $	an ^{-1}(1+x) = 	an ^{-1}(\frac{1}{1-x})$. Equating the arguments: $1+x = \frac{1}{1-x}$. This simplifies to: $(1+x)(1-x) = 1$. $1 - x^2 = 1$. $-x^2 = 0$,which implies $x = 0$.

The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is

Similar Questions

If $\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}$,then:

$\frac{d}{dx}[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)] = $

The value of $\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right)$ is

If $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$,then the value of $x$ is equal to

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module