Question

Prove that :
$\cot A-\tan A=\frac{2{\cos }^{2}A-1}{\sin A\cos A}$

Answer 1

$\cot A-\tan A$

$=\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}$

$=\frac{{\cos }^{2}A-{\sin }^{2}A}{\sin A\cos A}$

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

$=\frac{2{\cos }^{2}A-1}{\sin A\cos A}$