Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(1 + \text{sec A})}{\text{sec A}} = \frac{\text{sin² A}}{(\text{1 - cos A})}\) [Hint: Simplify LHS and RHS separately]

Answer 1

\(\frac{(1 + \text{sec A})}{\text{sec A}} = \frac{\text{sin² A}}{(\text{1 - cos A})}\)

L.H.S =\(\frac{(1 + \text{sec A})}{\text{sec A}} \)

\(\frac{(1 + \text{sec A})}{\text{sec A}} \)\(=\frac{ (1 + \frac{1}{\text{cos A}})}{(\frac{1}{\text{cos A}})}\)

\(= \frac{\frac{(\text{cos A + 1})}{\text{cosA}}}{(\frac{1}{\text{cos A}})}\)

\(= \frac{(\text{cosA + 1})}{\text{cos A}} × \frac{\text{cos A}}{1}\)

\(= \text{(1 + cos A)}\)

multiplying (1- cos A), in both denominator and numerator

\(⇒ \frac{\text{(1 - cos A)(1 + cos A)} }{\text{ (1 - cos A)}}\)

\(=\frac{ \text{(1 - cos² A)}}{\text{1 - cos A}} \)        [ Since, sin A + cos A = 1]

\(= \frac{\text{sin² A}}{\text{1 - cos A}}\)

= R. H.S