Question

Prove that tan^-1√x=1/2cos^-1[(1-x)÷(1+x)]

Answer 1

Step-by-step explanation:

$let \: \tan {}^{ - 1} ( \sqrt{x} ) = \alpha$

or

$x = \tan {}^{2} ( \alpha )$

RHS =

$\frac{1}{2} \cos {}^{ - 1} ( \frac{1 - x}{1 + x} )$

$\frac{1}{2} \cos {}^{ - 1} ( \frac{1 - \tan {}^{2} ( \alpha ) }{1 + \tan {}^{2} ( \alpha ) } )$

$\frac{1}{2} \cos {}^{ - 1} ( \frac{1 - \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } }{1 + \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } } )$

$\frac{1}{2} \cos {}^{ - 1} ( \frac{ \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) + \sin{}^{2} ( \alpha ) } )$

$\frac{1}{2} \cos {}^{ - 1} ( \frac{ \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) }{1 } )$

$\frac{1}{2} \cos {}^{ - 1} ( \cos( 2\alpha ) )$

$\frac{1}{2} \times 2 \alpha = \alpha$

$= \tan {}^{ - 1} ( \sqrt{x} )$

= LHS