Question : The value of $\frac{2 \cos ^3 \theta-\cos \theta}{\sin \theta-2 \sin ^3 \theta}$ is:
Option 1: $\sec \theta$
Option 2: $\sin \theta$
Option 3: $\cot \theta$
Option 4: $\tan \theta$
Correct Answer: $\cot \theta$
Solution :
$\frac{2 \cos ^3 \theta-\cos \theta}{\sin \theta-2 \sin ^3 \theta}$
$=\frac{\cos \theta (2 \cos^2 \theta - 1)}{\sin \theta (1 - 2 \sin^2 \theta)}$
We know that $\cos2 \theta = 1 - 2\sin^2 \theta= 2 \cos^2 \theta-1$,
Substituting these identities into the expression,
$=\frac{\cos \theta \cos2 \theta}{\sin \theta \cos2 \theta}$
$=\frac{\cos \theta}{\sin \theta} = \cot \theta$
Hence, the correct answer is $\cot \theta$.
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