Question

The value of $\frac{3\cos \theta +\cos 3\theta }{3\sin \theta -\sin 3\theta }$ is equal to:

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

Answer 1

Answer: (B)

${\cot }^3\theta$

we know that $\cos 3\theta =4{\cos }^3\theta -3\cos \theta ⟹\cos 3\theta +3\cos \theta =4{\cos }^3\theta$ -----(1)

and $\sin 3\theta =3\sin \theta -4{\sin }^3\theta ⟹3\sin \theta -\sin 3\theta =4{\sin }^3\theta$ ------(2)

Therefore, $\frac{3\cos \theta +\cos 3\theta }{3\sin \theta -\sin 3\theta }=\frac{4{\cos }^3\theta }{4{\sin }^3\theta }$

$=(\frac{\cos \theta }{\sin \theta }{)}^3=co{t}^3\theta$