if( ^ -1 x)^ 2 + ( ^ -1 x)^ 2 = 5 ^ 2 8 ,then find x. - Vidyadip

Question

$\text{if}(\tan ^{-1} x)^{2} + (\cot^{-1} x)^{2} = \frac{5{\pi}^{2}}{8} ,\text{then find x}.$

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Answer

$(\tan^{-1}\text{x})^{2} + (\cot^{-1}\text{x})^{2} = \frac{5{\pi}^{2}}{8} \Rightarrow (\tan^{-1}\text{x})^{2} + \bigg(\frac{\pi}{2} - \tan^{-1}\text{x}\bigg)^{2} = \frac{5{\pi}^{2}}{8}$
$\therefore 2 (\tan^{-1}\text{x})^{2} - \pi \tan^{-1}\text{x} - \frac{3{\pi}^{2}}{8} = 0$
$\tan^{-1}\text{x} = \frac{\pi\pm\sqrt{\pi^2+3\pi^{2}}}{4} = \frac{3\pi}{4}, \frac{-\pi}{4}$
$\Rightarrow \text{x} = -1$

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