Q4(iv):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ [Hint : Simplify LHS and RHS separately]

Given: The trigonometric identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$, where $A$ is an acute angle.

To Prove: The Left Hand Side (LHS) is equal to the Right Hand Side (RHS).

Step 1: Simplifying the Left Hand Side (LHS)

The expression is given as: $LHS = \frac{1 + \sec A}{\sec A}$

We know the fundamental trigonometric identity relating secant and cosine: $\sec A = \frac{1}{\cos A}$ [Definition of secant as the reciprocal of cosine].

Substituting this into the LHS expression:

$LHS = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$

To simplify the numerator, find a common denominator:

$LHS = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}$

Multiplying by the reciprocal of the denominator:

$LHS = \left( \frac{\cos A + 1}{\cos A} \right) \times \left( \frac{\cos A}{1} \right)$

Canceling the common term $\cos A$ in the numerator and denominator:

$LHS = 1 + \cos A$

Step 2: Simplifying the Right Hand Side (RHS)

The expression is given as: $RHS = \frac{\sin^2 A}{1 – \cos A}$

We use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$, which implies $\sin^2 A = 1 - \cos^2 A$ [Using the identity $\sin^2 \theta + \cos^2 \theta = 1$].

Substituting this into the RHS expression:

$RHS = \frac{1 - \cos^2 A}{1 - \cos A}$

Recognizing the numerator as a difference of squares, where $a^2 - b^2 = (a - b)(a + b)$. Here, $1 - \cos^2 A = (1)^2 - (\cos A)^2 = (1 - \cos A)(1 + \cos A)$:

$RHS = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}$

Canceling the common term $(1 - \cos A)$ in the numerator and denominator (assuming $1 - \cos A \neq 0$, which is true for acute angle $A \neq 0^\circ$):

$RHS = 1 + \cos A$

Step 3: Conclusion

Since $LHS = 1 + \cos A$ and $RHS = 1 + \cos A$, we have shown that $LHS = RHS$.

Final Answer: Since both sides simplify to $1 + \cos A$, the identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ is proven.