Question

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

Answer 1

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

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

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

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

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

Now, simplifying the RHS,

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

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

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

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

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

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

Therefore, $LHS=RHS={\tan }^{2}A$.