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Mathematics Question on Inverse Trigonometric Functions

The solution of Tan−1x+2cot−1x=2π3Tan^{-1}x+ 2cot^{-1}x=\frac {2\pi}{3} is

A

13\frac {1} {\sqrt{3}}

B

−13-\frac {1} {\sqrt{3}}

C

3\sqrt {3}

D

−3-\sqrt{3}

Answer

3\sqrt {3}

Explanation

Solution

We have, tan⁡−1x+2cot⁡−1x=2π3\tan^{-1} x + 2 \cot^{-1} x = \frac{2\pi}{3} ⇒tan⁡−1x+2tan⁡−11x=2π3\Rightarrow \tan^{-1} x +2 \tan^{-1} \frac{1}{x} = \frac{2\pi}{3} ⇒tan⁡−1x+tan⁡−1(2(1x)1−(1x)2)=2π3\Rightarrow \tan^{-1} x + \tan^{-1} \left(\frac{2\left(\frac{1}{x}\right)}{1- \left(\frac{1}{x}\right)^{2}}\right) = \frac{2\pi}{3} [∵2tan⁡−1x=tan⁡−12x1−x2]\left[\because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^{2}}\right] ⇒tan⁡−1x+tan⁡−1(2xx2−1)=2π3\Rightarrow \tan^{-1} x + \tan^{-1} \left(\frac{2x}{x^{2} -1} \right) = \frac{2\pi}{3} ⇒tan⁡−1(x+2xx2−11−2x2x2−1)=2π3\Rightarrow \tan^{-1} \left(\frac{x+ \frac{2x}{x^{2}-1}}{1- \frac{2x^{2}}{x^{2} - 1}}\right) = \frac{2\pi}{3} ⇒x3−x+2xx2−1−2x2=tan⁡(2π3)\Rightarrow \frac{x^{3} -x +2x}{x^{2} -1-2x^{2}} = \tan\left(\frac{2\pi}{3}\right) ⇒x3+x−1−x2=tan⁡(2π3)\Rightarrow \frac{x^{3} + x}{-1-x^{2}} =\tan \left(\frac{2\pi}{3}\right) ⇒x(x2+1)−1(x2+1)=−3x=3\Rightarrow \frac{x\left(x^{2} + 1\right)}{-1 \left(x^{2} +1\right)} = - \sqrt{3} x = \sqrt{3}