If tan ^{-1} a+tan ^{-1} b+tan ^{-1} c=pi,the | vedclass

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

Question

(C) We are given the equation: $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula for the sum of three inverse tangent functions: $	an ^

Vedclass · Accepted Answer

$a+b+c=a b c$

Vedclass · Answer

(C) We are given the equation: $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula for the sum of three inverse tangent functions: $	an ^{-1} x + 	an ^{-1} y + 	an ^{-1} z = 	an ^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} ight)$. Applying this to the given equation: $	an ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} ight) = \pi$. Taking the tangent of both sides: $\frac{a+b+c-abc}{1-ab-bc-ca} = 	an \pi$. Since $	an \pi = 0$,we have: $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$. This implies that the numerator must be zero: $a+b+c-abc = 0$. Therefore,$a+b+c = abc$.

If $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$,then which of the following is true?

Similar Questions

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

The value of $\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ is

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

$\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$ is equal to

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module