Question

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

• A. $1+{\cot }^2\theta$
• B. ${\cot }^4\theta$
• C. ${\cot }^3\theta$
• D. $2\cot \theta$

Answer 1

Answer: (C)

${\cot }^3\theta$

$\frac{3\cos \theta +\cos 3\theta }{3\sin \theta -\sin 3\theta }$
we know that $\sin 3x=3\sin x-4{\sin }^{3}x$ and $\cos 3x=4{\cos }^{3}x-3\cos x$
$=\frac{3\cos \theta +4{\cos }^3\theta -3\cos \theta }{3\sin \theta -3\sin \theta +4{\sin }^3\theta }$
$=\frac{4{\cos }^3\theta }{4{\sin }^3\theta }={\cot }^3\theta$