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