Question

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

Answer 1

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

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

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

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

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

Now simplify the RHS of $\left(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}\right)={\left(\frac{1-\tan A}{1-\cot A}\right)}^2$.

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

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

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

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