Question

Prove that:
${\left[\frac{1-\tan A}{1-\cot A}\right]}^2={\tan }^{2}A$

Answer 1

${\left(\frac{1-\tan A}{1-\cot A}\right)}^2$

$=\frac{{(1-\tan A)}^2}{{(1-\cot A)}^2}$

$=\frac{1+{\tan }^{2}A-2\tan A}{1+{\cot }^{2}A-2\cot A}$

$=\frac{{\sec }^{2}A-2\tan A}{{\csc }^{2}A-2\cot A}$

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

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

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

$={\tan }^{2}A$