\(\frac{(1 + \text{sec A})}{\text{sec A}} = \frac{\text{sin² A}}{(\text{1 - cos A})}\)
L.H.S =\(\frac{(1 + \text{sec A})}{\text{sec A}} \)
We begin with the Left Hand Side:
\(\frac{(1 + \text{sec A})}{\text{sec A}} \)\(=\frac{ (1 + \frac{1}{\text{cos A}})}{(\frac{1}{\text{cos A}})}\)
Substitute sec A with 1/cos A:
\(= \frac{\frac{(\text{cos A + 1})}{\text{cosA}}}{(\frac{1}{\text{cos A}})}\)
Simplify the complex fraction:
\(= \frac{(\text{cosA + 1})}{\text{cos A}} × \frac{\text{cos A}}{1}\)
Cancel out cos A:
\(= \text{(1 + cos A)}\)
To match the Right Hand Side, we multiply the numerator and denominator by (1 - cos A):
\(⇒ \frac{\text{(1 - cos A)(1 + cos A)} }{\text{ (1 - cos A)}}\)
Expand the numerator using the difference of squares formula (a² - b² = (a-b)(a+b)):
\(=\frac{ \text{(1 - cos² A)}}{\text{1 - cos A}} \) [ Using the identity sin² A + cos² A = 1, so sin² A = 1 - cos² A]
Substitute (1 - cos² A) with sin² A:
\(= \frac{\text{sin² A}}{\text{1 - cos A}}\)
= R. H. S