Question

Prove that \(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2\csc \theta\)

Hint:

Always use identities: \(\sin^2 \theta + \cos^2 \theta = 1\), and simplify numerator and denominator carefully.

Answer 1

To prove:
\[ \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta \]

Step 1: Find common denominator
\[ \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta} \]

Step 2: Expand numerator
\[ \sin^2 \theta + (1 + \cos \theta)^2 = \sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta \]
Using \(\sin^2 \theta + \cos^2 \theta = 1\),
\[ = 1 + 1 + 2 \cos \theta = 2 + 2 \cos \theta = 2(1 + \cos \theta) \]

Step 3: Substitute back in fraction
\[ \frac{2(1 + \cos \theta)}{(1 + \cos \theta) \sin \theta} = \frac{2}{\sin \theta} = 2 \csc \theta \]

Hence proved:
\[ \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta \]