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