Question

10 points || Inverse Trigonometry

if
$( \tan {}^{ - 1} (x) ){}^{2} + (\cot {}^{ - 1} (x) ) {}^{2} = \frac{5\pi {}^{2} }{8}$
, Find x​

Answer 1

SOLUTION ☺️

$= > ( \tan {}^{ - 1} x) {}^{2} + ( \cot {}^{ - 1} x) {}^{2} = \frac{5\pi {}^{2} }{8} \\ = > ( \tan {}^{ - 1} x + \cot {}^{ - 1} x) {}^{2} + 2 \tan {}^{ - 1}x \cot {}^{ - 1} x = \frac{5\pi {}^{2} }{8} \\ = > (\frac{\pi}{2} ) {}^{2} + 2 \tan {}^{ - 1} x(\pi - \tan {}^{ - 1} x) = \frac{5\pi {}^{2} }{8} \\ = > \frac{\pi {}^{2} }{4} + 2\pi \tan {}^{ - 1} x - 2( \tan {}^{ - 1} x) {}^{2} = \frac{5\pi {}^{2} }{8} \\ = > 2 \tan {}^{ - 1} x - 2( \tan {}^{ - 1} x) {}^{2} + \frac{\pi {}^{2} }{4} - \frac{5\pi {}^{2} }{8} = 0 \\ = > 2( \tan {}^{ - 1} x) {}^{2} - 2\pi \tan {}^{ - 1}x - \frac{\pi {}^{2} }{4} + \frac{5\pi {}^{2} }{8} = 0 \\ = > 2( \tan {}^{ - 1} x) {}^{2} - 2\pi \tan {}^{ - 1} x + \frac{5\pi {}^{2} - 2\pi {}^{2} }{8} = 0 \\ = > 2( \tan {}^{ - 1} x) {}^{2} - 2\pi \tan {}^{ - 1} x + \frac{3\pi {}^{2} }{8} = 0 \\ = > solving \: the \: quadratic \: equation \\ \\ = > \tan {}^{ - 1} x = \frac{2\pi + - \sqrt{4\pi {}^{2} - 4 \times 2 \times \frac{3\pi {}^{2} }{8} } }{2 \times 2 } \\ = > \tan {}^{ - 1}x = \frac{2\pi + - \sqrt{4 \pi {}^{2} -3\pi {}^{2} {} } }{4} \\ = > \tan {}^{ - 1} x = \frac{2\pi + - \pi}{4} \\ = > \tan {}^{ - 1} x = \frac{2\pi + \pi}{4} \: or \: \tan {}^{ - 1} x = \frac{2\pi - \pi}{4} \\ = > \tan {}^{ - 1} x = \frac{3\pi}{4} \: or \: \tan {}^{ - 1} = \frac{\pi}{4} \\ = > x = tan \frac{3\pi}{4} \: or \: x = tan \frac{\pi}{4} \\ = > x = 0 \: and \: 1$

hope it helps ✔️❣️

Answer 2

Answer in attachment!!!