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  4. The solution of Tan-1x+ 2cot-1x= (2π/3) is

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Q. The solution of $Tan^{-1}x+ 2cot^{-1}x=\frac {2\pi}{3}$ is

KCETKCET 2007Inverse Trigonometric Functions

Solution:

We have, $\tan^{-1} x + 2 \cot^{-1} x = \frac{2\pi}{3} $
$\Rightarrow \tan^{-1} x +2 \tan^{-1} \frac{1}{x} = \frac{2\pi}{3} $
$\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} $
$ \left[\because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^{2}}\right] $
$\Rightarrow \tan^{-1} x + \tan^{-1} \left(\frac{2x}{x^{2} -1} \right) = \frac{2\pi}{3} $
$\Rightarrow \tan^{-1} \left(\frac{x+ \frac{2x}{x^{2}-1}}{1- \frac{2x^{2}}{x^{2} - 1}}\right) = \frac{2\pi}{3} $
$\Rightarrow \frac{x^{3} -x +2x}{x^{2} -1-2x^{2}} = \tan\left(\frac{2\pi}{3}\right) $
$\Rightarrow \frac{x^{3} + x}{-1-x^{2}} =\tan \left(\frac{2\pi}{3}\right) $
$\Rightarrow \frac{x\left(x^{2} + 1\right)}{-1 \left(x^{2} +1\right)} = - \sqrt{3} x = \sqrt{3}$