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$\large\frac{3\cos\theta+\cos 3\theta}{3\sin \theta-\sin 3\theta}$ is equal to

$\begin{array}{1 1}(a)\;1+\cot^2\theta&(b)\;\cot^4\theta\\(c)\;\cot^3\theta&(d)\;2\cot\theta\end{array}$

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1 Answer

$\large\frac{3\cos\theta+\cos 3\theta}{3\sin\theta-\sin 3\theta}=\frac{4\cos^3\theta}{4\sin^3\theta}$
$\qquad\qquad\qquad=\large\frac{\cos^3\theta}{\sin^3\theta}$
$\qquad\qquad\qquad=\cot^3\theta$
Hence (c) is the correct option.
answered Oct 9, 2013 by sreemathi.v
 

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