Question

$If\sin \left[{\cot }^{-1}\left(x+1\right)\right]=\cos \left({\tan }^{-1}x\right),$ then find $x$

Answer 1

$sin\left[co{t}^{-}1\right(x+1\left)\right]=cos(\frac{\pi }{2}-co{t}^{-}1(x+1\left)\right)=cos\left(ta{n}^{-}1\right(x\left)\right)$

if $cosx=cos\theta$

then $x=2n\pi +\theta ,x=2n\pi -\theta$

$\frac{\pi }{2}-co{t}^{-}1(x+1)=2n\pi +ta{n}^{-}1\left(x\right)$

apply tan on both sides

$cot\left(co{t}^{-}1\right(x+1\left)\right)=tan(2n\pi +ta{n}^{-}1(x\left)\right)$

$x+1=x$

x does not exist

now consider another case

$x=2n\pi -\theta$

$\frac{\pi }{2}-co{t}^{-}1(x+1)=2n\pi -ta{n}^{-}1\left(x\right)$

apply tan on both sides

$cot\left(co{t}^{-}1\right(x+1\left)\right)=tan(2n\pi -ta{n}^{-}1(x\left)\right)$

$x+1=-x$

$x=\frac{-1}{2}$