Question

Prove that : \(\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A\)

Hint:

When proving trig identities, keep an eye on the RHS to decide whether to convert everything to \(\sin/\cos\) early or use identities like \(1 - \sec^2 A = -\tan^2 A\).

Answer 1

Step 1: Solving OR (B):
1. Take \(\tan A\) common: \(\tan A \left[ \frac{1}{1 + \sec A} - \frac{1}{1 - \sec A} \right]\). 2. Take LCM: \(\tan A \left[ \frac{(1 - \sec A) - (1 + \sec A)}{1 - \sec^2 A} \right]\). 3. Simplify numerator and denominator: \(\tan A \left[ \frac{-2 \sec A}{-\tan^2 A} \right]\). 4. Simplify: \(\frac{2 \sec A}{\tan A} = 2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2 \csc A\).
Step 2: Final Answer (OR):
LHS = RHS. Hence proved.