Question

Prove that
\[ \left( \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} \right) \left( \frac{\sin A}{1-\cos A} + \frac{1-\cos A}{\sin A} \right) = 4 \csc A \cot A. \]

Hint:

Trigonometric Proofs: Use trigonometric identities and simplifications.

Answer 1

Consider the first term:
\[ \frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} \] Taking LCM:
\[ = \frac{\sin^2 A + (1+\cos A)^2}{\sin A (1+\cos A)} \] Expanding:
\[ = \frac{\sin^2 A + 1 + 2\cos A + \cos^2 A}{\sin A (1+\cos A)} \] \[ = \frac{2 + 2\cos A}{\sin A (1+\cos A)} \] \[ = \frac{2(1+\cos A)}{\sin A (1+\cos A)} = \frac{2}{\sin A} \] Similarly,
\[ \frac{\sin A}{1-\cos A} + \frac{1-\cos A}{\sin A} \] Solving similarly, we get:
\[ = \frac{2}{\sin A} \] Multiplying both terms:
\[ \frac{2}{\sin A} \times \frac{2}{\sin A} = \frac{4}{\sin^2 A} \] \[ = 4 \csc A \cot A \] Thus, the identity is proved.
Correct Answer: \( \frac{5}{12} \)