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 $x$ which satisfies $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$ is

If $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$,then $\frac{dy}{dx} = $

If $\tan^{-1} \frac{a+x}{a} + \tan^{-1} \frac{a-x}{a} = \frac{\pi}{6}$,then $x^2 =$

The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

If $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module