Question

Prove that coses A - cot A =√1-cos A÷√1+cos A

Answer 1

$\huge\mathfrak\red{Answer}$

$\csc \alpha - \cot \alpha \\ \frac{1}{ \sin \alpha } - \frac{ \cos \alpha }{ \sin \alpha } \\ \\ take \: lcm \\ \\ \frac{1 - \cos \alpha }{ \sin \alpha } \\ \\ \frac{ {( \sqrt{1 - \cos \alpha )} }^{2} }{ \sqrt{ (1 - {cos}^{2} \alpha ) } } \\ \\ \frac{ \sqrt{ {(1 - \cos \alpha )}^{2} } }{ \sqrt{(1 - \cos \alpha )(1 + \cos \alpha ) } } \\ \\ \frac{ \sqrt{1 - \cos \alpha } }{ \sqrt{1 + \cos \alpha } } \\ |hence \: proved|$

Answer 2

$\huge\mathfrak\red{Answer}$

$\csc \alpha - \cot \alpha = \frac{ \sqrt{1 - \cos \alpha } }{ \sqrt{1 + \cos \alpha } } \\ \\ \frac{1}{ \sin \alpha } - \frac{ \cos \alpha }{ \sin \alpha } = \frac{ \sqrt{1 - \cos \alpha } }{ \sqrt{1 + \cos \alpha } } \\ \\ multiply \:and \: divide \: by \sqrt{1 - \cos \alpha } \\ on \: right \: hand \: side \\ \\ \frac{1 - \cos \alpha }{ \sin \alpha } = \frac{ \sqrt{ {(1 - \cos \alpha ) }^{2} } }{ \sqrt{1 - {cos}^{2} \alpha } } \\ \\ \frac{1 - \cos \alpha }{ \sin \alpha } = \frac{1 - cos \alpha }{ \sqrt{ {sin}^{2} \alpha } } \\ \\ \frac{1 - cos \alpha }{ \sin \alpha } = \frac{1 - \cos \alpha }{ \sin \alpha }$