Question

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

Answer 1

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

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

$=\frac{{\sec }^{2}A}{cose{c}^{2}A}$ $=\frac{1+{\tan }^{2}A-2\tan A}{1+{\cot }^{2}A-2\cot A}$

$=\frac{1}{{\cos }^{2}A}\times \frac{{\sin }^{2}A}{1}$ $=\frac{{\sec }^{2}A-2\tan A}{cose{c}^{2}A-2\cot A}$

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

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

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

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