Question

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

Hint:

Be very careful with signs when applying identities like \(1 - \sec^2 \theta\). Since \(\sec^2 \theta = 1 + \tan^2 \theta\), then \(1 - \sec^2 \theta = -\tan^2 \theta\). Missing a negative sign is a common error here.

Answer 1

Step 1: Understanding the Concept:
Simplifying complex trigonometric fractions often involves finding a common denominator and using fundamental identities like \(1 - \sec^2 A = -\tan^2 A\).
Step 2: Key Formula or Approach:
\[ \sec^2 A - \tan^2 A = 1 \implies 1 - \sec^2 A = -\tan^2 A \]
Step 3: Detailed Explanation:
LHS = \(\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A}\)
Factor out \(\tan A\):
\[ = \tan A \left[ \frac{1}{1 + \sec A} - \frac{1}{1 - \sec A} \right] \]
Taking common denominator:
\[ = \tan A \left[ \frac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A)(1 - \sec A)} \right] \]
\[ = \tan A \left[ \frac{1 - \sec A - 1 - \sec A}{1 - \sec^2 A} \right] \]
\[ = \tan A \left[ \frac{-2 \sec A}{-\tan^2 A} \right] \]
\[ = \frac{2 \tan A \sec A}{\tan^2 A} = \frac{2 \sec A}{\tan A} \]
Convert to sine and cosine:
\[ = \frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A = \text{RHS} \]
Step 4: Final Answer:
Hence proved.