First slide

Multiple and sub- multiple Angles

Question

If $\frac{\tan 3 A}{\tan A} = k$ , then $\frac{\sin 3 A}{\sin A}$ is equal to

Moderate

A

$\frac{2 k}{k - 1}, k \in R$
B

$\frac{2 k}{k - 1}, k \in [1 / 3, 3]$
C

$\frac{2 k}{k - 1}, k \notin [1 / 3, 3]$
D

$\frac{k - 1}{2 k}, k \notin [1 / 3, 3]$

Solution

We have, $\frac{\tan 3 A}{\tan A} = k \Rightarrow \frac{3 - \tan^{2} A}{1 - 3 \tan^{2} A} = k \Rightarrow \tan^{2} A = \frac{k - 3}{3 k - 1}$
Now,
$\begin{array}{l} \frac{\sin 3 A}{\sin A} = 3 - 4 \sin^{2} A = 3 - \frac{4}{1 + \cot^{2} A} \\ = 3 - \frac{4}{1 + \frac{3 k - 1}{k - 3}} = \frac{2 k}{k - 1} \end{array}$
Again, $\frac{\sin 3 A}{\sin A} = 3 - 4 \sin^{2} A$
$\begin{array}{l} \Rightarrow \frac{2 k}{k - 1} = 3 - 4 \sin^{2} A \\ \Rightarrow 4 \sin^{2} A = 3 - \frac{2 k}{k - 1} \\ \Rightarrow \sin^{2} A = \frac{k - 3}{4 (k - 1)} \\ \Rightarrow 0 \leq \frac{k - 3}{4 (k - 1)} \leq 1 [∵ 0 \leq \sin^{2} A \leq 1] \end{array}$
$\Rightarrow k < \frac{1}{3}$ or $k > 3$
Hence, $\frac{\sin 3 A}{\sin A} = \frac{2 k}{k - 1}$ , where $k < \frac{1}{3}$ or, $k > 3$ .

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