If tan^{-1}x + tan^{-1}y + tan^{-1}z = pi,the | vedclass

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

Question

(A) Given the equation: $	an^{-1}x + 	an^{-1}y + 	an^{-1}z = \pi$. Using the formula for the sum of three inverse tangents: $	an^{-1}\left( \frac{

Vedclass · Accepted Answer

$xyz$

Vedclass · Answer

(A) Given the equation: $	an^{-1}x + 	an^{-1}y + 	an^{-1}z = \pi$. Using the formula for the sum of three inverse tangents: $	an^{-1}\left( \frac{x+y+z-xyz}{1-xy-yz-zx} ight) = \pi$. Taking the tangent of both sides: $\frac{x+y+z-xyz}{1-xy-yz-zx} = 	an(\pi)$. Since $	an(\pi) = 0$,we have: $\frac{x+y+z-xyz}{1-xy-yz-zx} = 0$. This implies the numerator must be zero: $x+y+z-xyz = 0$. Therefore: $x+y+z = xyz$.

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$,then $x + y + z$ is equal to

Similar Questions

If $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$,then $\alpha=$

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is equal to

The value of $\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is

If $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$,then $\frac{dy}{dx}$ equals-

If $y = \sin^{-1}(x\sqrt{1 - x} + \sqrt{x}\sqrt{1 - x^2})$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module