Question

tan^-1(1+x/1-x)= π\4+ tan^-1 x...pls provide that​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{tan^{ - 1} \left( \dfrac{1 + x}{1 - x}\right) }$

$\sf \: Put \: x = tan(θ) \implies\sf{tan^{-1}(x)=\theta}$

So,

$\sf{ = tan^{ - 1} \left \{ \dfrac{1 + tan( \theta)}{1 - tan( \theta)}\right \} }$

$\sf{ = tan^{ - 1} \left \{ \dfrac{tan \left( \dfrac{\pi}{4} \right) + tan( \theta)}{1 - tan \left( \dfrac{\pi}{4} \right) \cdot tan( \theta)}\right \}} \\ \\ \sf{ = tan^{ - 1} \left \{ tan \left( \dfrac{\pi}{4} + \theta\right)\right \}}$

$\sf{ = \dfrac{\pi}{4} + \theta }$

$\sf{ = \dfrac{\pi}{4} + tan^{ - 1}( x )}$