Question : The value of $\frac{\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}$ is equal to:
Option 1: $\sin\theta$
Option 2: $\cos\theta$
Option 3: $\tan\theta$
Option 4: $\cot\theta$
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Correct Answer: $\tan\theta$
Solution :
Given: $\frac{\sin\theta–2\sin^{3}\theta}{2\cos^{3}\theta–\cos\theta}$
$=\frac{\sin\theta(1–2\sin^{2}\theta)}{\cos\theta(2\cos^{2}\theta–1)}$
We know that $\cos2\theta=1-2\sin^{2}\theta=2\cos^{2}\theta-1$
So, $\frac{\sin\theta(\cos2\theta)}{\cos\theta(\cos2\theta)}$
$=\frac{\sin\theta}{\cos\theta}$
$=\tan\theta$
Hence, the correct answer is $\tan\theta$.
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