Question

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

Answer 1

we have,

$L.H.S.$

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

$=\frac{{\sec }^{2}A}{{\csc }^{2}A}$

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

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

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

$R.H.S.$

Hence proved.

Again,

We have

$LHS.$

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

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

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

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

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

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

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

$RHS.$

Hence proved.