Question

tignometry questions of class 12 ​

Answer 1

Answer:

We know

$\tan {}^{ - 1} (x) + \cot {}^{ - 1} (x) = \frac{\pi}{2} \\ \\ \cot {}^{ - 1} (x) = \frac{\pi}{2} - \tan { }^{ - 1} (x) \\ \\ let \: \: \: \tan { }^{ - 1} (x) = y \\ \\ \cot { }^{ - 1} (x) = \frac{\pi}{2} - y$

Now put values in equation

${y}^{2} + ( \frac{\pi}{2} - y) {}^{2} = \frac{5 {\pi}^{2} }{8} \\ \\ {y}^{2} + {y}^{2} + \frac{ {\pi}^{2} }{4} - \pi \: y = \frac{5 {\pi}^{2} }{8} \\ \\ 16 {y}^{2} + 2 {\pi}^{2} - 8\pi \: y = 5 {\pi}^{2} \\ \\ 16 {y}^{2} - 8\pi \: y - 3 {\pi}^{2} = 0 \\ \\ y = \frac{8\pi \pm \sqrt{64 {\pi}^{2} + 4(16)(3 {\pi}^{2}) } }{32} \\ \\ y = { \frac{8\pi \pm16\pi}{32} } = \frac{24\pi}{32} \: \: and \: \: \frac{ - 8\pi}{32} \\ \\ \tan {}^{1} (x) = \frac{3\pi}{4} \: \: and \: \: \frac{ - \pi}{4} \\ \\ x = \tan( \frac{3\pi}{4} ) \: and \: \: { \tan( \frac{ - \pi}{4} ) } \\ \\ x = - 1$