# Get Answers to all your Questions

#### If $3\tan^{-1}x+\cot^{-1}x=\pi$, then x equals toA. 0B. 1C. -1D. 1/2

Given That, $3 \tan ^{-1}x+\cot^{-1}x=\pi$

$\Rightarrow 2 \tan^{-1}x+\tan^{-1}x+\cot^{-1}x=\pi$

$\Rightarrow 2 \tan^{-1}x=\pi-\frac{\pi}{2}$  $\left [ Scince, \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} \right ]$

$\Rightarrow \tan^{-1}\frac{2x}{1-x^{2}}=\frac{\pi}{2}$ $\left [ Scince, 2tan^{-1}x=\tan^{-1}\frac{2x}{1-x^{2}} \right ]$

$\Rightarrow \frac{2x}{1-x^{2}}=\tan \frac{\pi}{2}$

$\Rightarrow \frac{2x}{1-x^{2}}=\tan \frac{1}{0}$

Cross multiplying

$\Rightarrow 1-x^{2}=0$

$\Rightarrow x^{2}=\pm 1$

Here, only x = 1 satisfies the given equation.

Note: By putting x=-1 in the given equation, we get,

$3 \tan^{-1}(-1)+cot^{-1}(-1)=\pi$

$3 \tan^{-1} \left [ \tan\left (\frac{-\pi}{4} \right )\right ]+cot^{-1} \left [ \cot \left (\frac{-\pi}{4} \right )\right ]=\pi$

$3 \tan^{-1} \left [- \tan\left (\frac{\pi}{4} \right )\right ]+cot^{-1} \left [- \cot \left (\frac{\pi}{4} \right )\right ]=\pi$

$3 \tan^{-1} \left [\tan\left (\frac{\pi}{4} \right )\right ]+\pi-cot^{-1} \left [\cot \left (\frac{\pi}{4} \right )\right ]=\pi$

$-3\times \frac{\pi}{4}+\pi-\frac{\pi}{4}=\pi$

$-\pi+\pi=\pi$

$0\neq \pi$

Hence, x = -1 does not satisfy the given equation.