Question

If the value of $\frac{\cot \theta}{\cot 3\theta}=k$, where $\theta \neq \frac{n\pi}{6}, n \in I$, then the value of $\frac{\csc \theta}{\csc 3\theta}$ is equal to

• A. $\frac{2k}{k-1}$

Answer 1

Answer: (A)

$\frac{2k}{k-1}$

Answer 2

Given $\frac{\cot\theta}{\cot3\theta}=k$, write it as $\frac{\cos\theta\sin3\theta}{\sin\theta\cos3\theta}=k$ so that $\frac{\sin3\theta}{\sin\theta}= k\,\frac{\cos3\theta}{\cos\theta}$. Using triple angle formulas, we write

$$\frac{\sin3\theta}{\sin\theta}=4\cos^2\theta-1\quad \text{and}\quad \frac{\cos3\theta}{\cos\theta}=4\cos^2\theta-3.$$

Set

$$4\cos^2\theta-1=k(4\cos^2\theta-3),$$

solve for $\cos^2\theta$ and substitute back to find:

$$\frac{\sin3\theta}{\sin\theta}=\frac{2k}{k-1}.$$

Since $\frac{\csc\theta}{\csc3\theta}=\frac{\sin3\theta}{\sin\theta}$, the answer is $\frac{2k}{k-1}$.