Question

Prove that:

$\left(\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}\right)=(\frac{1-tanA}{cotA}{)}^2=ta{n}^{2}A$

Answer 1

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

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

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