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

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