Question

$solve \: it \\ \\ \tan^{ - 1} \frac{1}{x} + \tan^{ - 1} 2 = \frac{\pi}{3} \\ \\ find \: the \: value \: of \: x \:$

Chapter: Inverse Trigonometry

Answer: 2

Please don't give useless answer. It's a Request.

PLEASE SOLVE IT .

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ Now, \: \: {tan}^{ - 1} \frac{1}{x} + {tan}^{ - 1} 2 = \frac{\pi}{2} \\ \\ Or, \: \: {tan}^{ - 1} ( \frac{ \frac{1}{x} + 2}{1 - ( \frac{1}{x} \times 2)} ) = \frac{\pi}{2} \\ \\ Or, \: \: \frac{1 + 2x}{x - 2} = tan \frac{\pi}{2} \\ \\ Or, \: \: \frac{1 + 2x}{x - 2} = \frac{1}{cot \frac{\pi}{2} } \\ \\ Or, \: \: \frac{1 + 2x}{x - 2} = \frac{1}{ 0} \\ \\ Or, \: \: x - 2 = 0 \\ \\ So, \: \: x = 2 \\ \\ Thank \: \: you \: \: for \: \: your \: \: question.$

Answer 2

see attachment.........hope it will help u