Question

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

Answer 1

$L.H.S=\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}$

$=\frac{1+ta{n}^{2}A}{1+\frac{1}{ta{n}^{2}A}}$

$=\frac{(1+ta{n}^{2}A)ta{n}^{2}A}{(ta{n}^{2}A+1)}$

$\therefore \frac{1+ta{n}^{2}A}{1+co{t}^{2}A}=ta{n}^{2}A$