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If 3\tan^{-1}x+\cot^{-1}x=\pi, then x equals to

A. 0

B. 1

C. -1

D. 1/2

Answers (1)

Answer: B

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


0\neq \pi

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




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