Question

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

Hint:

The expression \(\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}\) always simplifies to \(\tan^2 \theta\). Memorizing this can save you significant time in exams.

Answer 1

Step 1: Understanding the Concept:
To prove trigonometric identities, it is often helpful to convert all terms into \(\sin\) and \(\cos\).
Step 2: Key Formula or Approach:
\(\tan A = \frac{\sin A}{\cos A}\) and \(\sec A = \frac{1}{\cos A}\).
Step 3: Detailed Explanation:
1. Take LHS: \(\frac{\sin A - \frac{\sin A}{\cos A}}{\sin A + \frac{\sin A}{\cos A}}\)
2. Take \(\sin A\) common from numerator and denominator:
\[ \frac{\sin A (1 - \frac{1}{\cos A})}{\sin A (1 + \frac{1}{\cos A})} \] 3. Cancel \(\sin A\):
\[ \frac{1 - \frac{1}{\cos A}}{1 + \frac{1}{\cos A}} \] 4. Substitute \(\frac{1}{\cos A} = \sec A\):
\[ \frac{1 - \sec A}{1 + \sec A} = \text{RHS} \]
Step 4: Final Answer:
LHS = RHS. Hence proved.