Question

Prove that \(\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \frac{1}{\sec A + \tan A}\).

Hint:

Rationalization is the key to solving trigonometric identities involving square roots.
Multiplying the numerator and denominator by the conjugate of the denominator helps eliminate the root easily!

Answer 1

Correct Answer: 4

Step 1: Write the LHS.
\(\text{LHS} = \sqrt{\frac{1 - \sin A}{1 + \sin A}}\).
Step 2: Rationalise the expression under the root.
Multiply numerator and denominator inside the root by \((1 - \sin A)\): \(\text{LHS} = \sqrt{\frac{(1-\sin A)^2}{(1+\sin A)(1-\sin A)}} = \sqrt{\frac{(1-\sin A)^2}{1 - \sin^2 A}}\).
Step 3: Use the Pythagorean identity.
Since \(1 - \sin^2 A = \cos^2 A\): \(\text{LHS} = \sqrt{\frac{(1-\sin A)^2}{\cos^2 A}} = \frac{1 - \sin A}{\cos A}\) (taking positive square root).
Step 4: Split the fraction.
\(\text{LHS} = \frac{1}{\cos A} - \frac{\sin A}{\cos A} = \sec A - \tan A\).
Step 5: Simplify the RHS.
\(\text{RHS} = \frac{1}{\sec A + \tan A}\). Multiply numerator and denominator by \((\sec A - \tan A)\): \(= \frac{\sec A - \tan A}{\sec^2 A - \tan^2 A} = \frac{\sec A - \tan A}{1} = \sec A - \tan A\). (Using identity \(\sec^2 A - \tan^2 A = 1\).)
Step 6: LHS = RHS. Hence proved.
\[ \boxed{\sqrt{\dfrac{1-\sin A}{1+\sin A}} = \dfrac{1}{\sec A + \tan A}} \]