2. inverse trigonometric function — ગણિત ધોરણ 12 સાયન્સ — Question

==> ${({\tan ^{ - 1}}x + {\cot ^{ - 1}}x)^2} - 2{\tan ^{ - 1}}x\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}x} \right) = \frac{{5{\pi ^2}}}{8}$

==> $\frac{{{\pi ^2}}}{4} - 2 \times \frac{\pi }{2}{\tan ^{ - 1}}x + 2{({\tan ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$

==> $2{({\tan ^{ - 1}}x)^2} - \pi {\tan ^{ - 1}}x - \frac{{3{\pi ^2}}}{8} = 0$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4},\frac{{3\pi }}{4}$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4} \Rightarrow x = - 1$.