If tan ^{-1}x + tan ^{-1}y + tan ^{-1}z = fra | vedclass

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

Question

(D) Given that $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = \frac{\pi }{2}.$Using the formula $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = 	an ^{-1}\left

Vedclass · Accepted Answer

$xy + yz + zx - 1 = 0$

Vedclass · Answer

(D) Given that $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = \frac{\pi }{2}.$Using the formula $	an ^{-1}x + 	an ^{-1}y + 	an ^{-1}z = 	an ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} ight),$$	an ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} ight) = \frac{\pi }{2}.$Taking tangent on both sides:$\frac{x + y + z - xyz}{1 - xy - yz - zx} = 	an \left( \frac{\pi }{2} ight) = \infty.$For the expression to be undefined (approaching $\infty$),the denominator must be zero:$1 - xy - yz - zx = 0.$Rearranging the terms,we get:$xy + yz + zx = 1$ or $xy + yz + zx - 1 = 0.$Thus,option $(d)$ is correct.

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

Similar Questions

If $\sec^{-1} x = \csc^{-1} y$,then $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

If $\operatorname{Tan}^{-1} \frac{1}{3}+\operatorname{Tan}^{-1} \frac{1}{7}+\operatorname{Tan}^{-1} \frac{1}{13}+\ldots+\operatorname{Tan}^{-1} \frac{1}{n^2+n+1}=\operatorname{Tan}^{-1} \theta$,then $\theta=$

If $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} \right) = 2a$, such that $y \frac{dy}{dx} = x \cdot f(a)$, then the value of $f \left( \frac{2\pi}{3} \right)$ is

$\sin^{-1} \sqrt{\frac{x}{x+a}}$ 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