Question

solve this question..
don't copy from internet please​

Answer 1

$\huge\underline\bold\red{★Answer★}$

$2 {tan}^{ - 1} (cosx) = {tan}^{ - 1} \frac{2cosx}{1 - {cos}^{2} x} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {tan}^{ - 1} \frac{2cosx}{ {sin}^{2}x } \\ now \: \: 2 {tan}^{ - 1} (cosx) = {tan}^{ - 1} (2cosex) \\ \implies {tan}^{ - 1} \frac{2cosx}{ {sin}^{2}x} = {tan}^{ - 1} (2cosecx) \\ \implies \frac{2cosx }{ {sin}^{2} x} = 2cosecx \\ \implies \frac{cosx}{ {sin}^{2} x} = \frac{1}{sinx} \\ \implies \frac{cosx}{sinx} = 1 \\ \implies cotx = 1 \\ \implies x = \frac{\pi}{4}$