Question

y=tan inverse (1-x/1+x)​

Answer 1

Answer:

Your answer attached in the photo

Answer 2

Heya!!

$\sf \: y = {tan}^{ (- 1)}( \frac{1 - x}{1 + x} )$

$\red { \sf \: put \: x \: = tan \theta}$

$\therefore \sf \: y = {tan}^{ (- 1)} ( \frac{1 - tan \theta}{1 + tan \theta} )$

$\longrightarrow \sf {tan}^{( - 1)} ( \frac{tan \frac{\pi}{4} - tan \theta}{1 + tan \frac{\pi}{4} (tan \theta)} )$

$\longrightarrow \sf \: \cancel {tan}^{( - 1)} \cancel {tan}( \frac{\pi}{4} - \theta )$

$\longrightarrow \sf \frac{\pi}{4} - \theta$

$\therefore \sf y = \frac{\pi}{4} - tan ^{( - 1)} x$

$\boxed{ \sf \frac{dy}{dx} = - \frac{1}{1 + {x}^{2} } }$