Question

If (tan-1 x )2 + (cot-1 x)2 =5pi^ / 8, then find x.

Answer 1

Answer:

-1

Step-by-step explanation:

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

Since, $cot^{-1}x = \frac{\pi}{2}-tan^{-1}x$

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

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

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

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

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

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

By quadratic formula,

$tan^{-1}x=\frac{\pi\pm \sqrt{\pi^2+3\pi^2}}{4}$

$tan^{-1}x=\frac{\pi\pm \sqrt{4\pi^2}}{4}$

$tan^{-1}x=\frac{\pi \pm 2\pi}{4}$

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

$\implies x = -1$

Answer 2

Answer:

the answer is given in the picture attached