Question

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

Answer 1

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

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

$=\left(\frac{{\sec }^{2}A}{{\text{cosec}}^{2}A}\right)=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A\text{(Proved)}$

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

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

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

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

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

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

$={\tan }^{2}A$ $\text{( (proved)}$