Question

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

Answer 1

LHS$=\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}=\frac{se{c}^{2}A}{cose{c}^{2}A}=\frac{si{n}^{2}A}{co{s}^{2}A}=ta{n}^{2}A=RHS$