Question

Prove that : $\frac{2 \cos^3} \theta - \cos \theta{\sin \theta - 2 \sin^3 \theta} = \cot \theta$

Hint:

Notice that both $(2 \cos^2 \theta - 1)$ and $(1 - 2 \sin^2 \theta)$ are actually formulas for $\cos 2\theta$.

Answer 1

Step 1: Write the given expression.
We need to prove the identity:
\(\frac{2\cos^3\theta - \cos\theta}{\sin\theta - 2\sin^3\theta} = \cot\theta\)

Step 2: Simplify the numerator.
Take \(\cos\theta\) common from the numerator:
\(2\cos^3\theta - \cos\theta = \cos\theta(2\cos^2\theta - 1)\)

Step 3: Simplify the denominator.
Take \(\sin\theta\) common from the denominator:
\(\sin\theta - 2\sin^3\theta = \sin\theta(1 - 2\sin^2\theta)\)

Step 4: Use the trigonometric identity.
We know the identity:
\(2\cos^2\theta - 1 = 1 - 2\sin^2\theta\)

Substitute this in the expression:
\[ \frac{\cos\theta(2\cos^2\theta - 1)}{\sin\theta(1 - 2\sin^2\theta)} = \frac{\cos\theta(1 - 2\sin^2\theta)}{\sin\theta(1 - 2\sin^2\theta)} \]

Step 5: Cancel the common term.
The term \((1 - 2\sin^2\theta)\) cancels from numerator and denominator:
\[ \frac{\cos\theta}{\sin\theta} \]

Step 6: Express in terms of cotangent.
\[ \frac{\cos\theta}{\sin\theta} = \cot\theta \]

Final Result:
Hence, the given identity is proved:
\(\frac{2\cos^3\theta - \cos\theta}{\sin\theta - 2\sin^3\theta} = \cot\theta\).