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Q. $\sin (\cot^{-1}(\tan\,\cos^{-1}x)$ is equal to

Inverse Trigonometric Functions

Solution:

Put $cos^{-1} x= \theta$
$ \therefore x= cos \,\theta $
$\therefore tan\, \theta = \frac{\sqrt{1-x^{2}}}{x} $
image
$ \therefore tan\left(cos^{-1} x\right) = tan\, \theta = \frac{\sqrt{1-x^{2}}}{x}$
$ \therefore cot^{-1} \left(tan\left(cos^{-1}\right)\right) = cot^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right) = \phi$ (say)
image
$ \therefore cot\, \phi = \frac{\sqrt{1-x^{2}}}{x} $
$ \therefore sin\, \phi = \frac{x}{1} $
$ \therefore sin(cot^{-1}\left(tan\left(cos^{-1}\,x\right)\right)$
$ = sin \,\phi = x$