Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:$\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 \- cot A}})^²= \text{tan² A}$

Answer 1

$\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 \- cot A}})^²= \text{tan² A}$

LHS, $\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})}$

$= \frac{\text{sec² A}}{\text{cosec² A}}$

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

$= (\frac{1}{\text{cos² A}}) × (\frac{\text{sin² A}}{1})$

$=\text{ tan² A}$
= RHS