Question

Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

Hint:

Factor cubic terms and use \(\sin^2 \theta + \cos^2 \theta = 1\) identity to simplify expressions.

Answer 1

To prove:
\[\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0\]

Step 1: Factorize numerator and denominator
\[\cos \theta - 2 \cos^3 \theta = \cos \theta (1 - 2 \cos^2 \theta)\] \[\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta)\]

Step 2: Rewrite the expression
\[\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} + \cot \theta\] Recall \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), so rewrite as:
\[\frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} + \frac{\cos \theta}{\sin \theta}\]

Step 3: Find a common denominator
\[= \frac{\cos \theta (1 - 2 \cos^2 \theta) + \cos \theta (1 - 2 \sin^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)}\]

Step 4: Simplify the numerator
\[\cos \theta \left[(1 - 2 \cos^2 \theta) + (1 - 2 \sin^2 \theta)\right] = \cos \theta [2 - 2(\cos^2 \theta + \sin^2 \theta)]\] Since \(\cos^2 \theta + \sin^2 \theta = 1\),
\[= \cos \theta [2 - 2 \times 1] = \cos \theta \times 0 = 0\]

Step 5: Final expression
\[\frac{0}{\sin \theta (1 - 2 \sin^2 \theta)} = 0\]

Hence proved:
\[\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0\]