Q4(vi):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

Given: An identity involving an acute angle $A$, specifically $\sqrt{\frac{1 + \sin A}{1 – \sin A}}$.

To Prove: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

Step 1: Consider the Left Hand Side (LHS) of the identity.

LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

Step 2: Rationalize the denominator inside the square root.

To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 + \sin A)$.

LHS = $\sqrt{\frac{(1 + \sin A) \times (1 + \sin A)}{(1 - \sin A) \times (1 + \sin A)}}$

Step 3: Simplify the expression using algebraic identities.

In the numerator, we have $(1 + \sin A)(1 + \sin A) = (1 + \sin A)^2$.

In the denominator, we use the difference of squares identity: $(a - b)(a + b) = a^2 - b^2$.

Here, $(1 - \sin A)(1 + \sin A) = 1^2 - \sin^2 A = 1 - \sin^2 A$.

LHS = $\sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$

Step 4: Apply the fundamental trigonometric identity.

We know that $\sin^2 A + \cos^2 A = 1$, which implies that $1 - \sin^2 A = \cos^2 A$.

Substituting this into our expression:

LHS = $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$

Step 5: Extract the square root.

Since the square root of a squared term is the term itself (given $A$ is an acute angle, $\cos A > 0$ and $1 + \sin A > 0$):

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

Step 6: Separate the terms in the fraction.

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

Step 7: Apply trigonometric reciprocal and quotient identities.

We know that $\frac{1}{\cos A} = \sec A$ and $\frac{\sin A}{\cos A} = \tan A$.

LHS = $\sec A + \tan A$

Conclusion:

Since the Left Hand Side (LHS) is equal to the Right Hand Side (RHS), the identity is proven.

Final Answer: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$ is proved.