Question

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

• A. $\frac{3k}{k-1},k\in R$
• B. $\frac{2k}{k+1},k\in (\frac{1}{3},3)$
• C. $\frac{2k}{k-1},k\notin (\frac{1}{3},3)$
• D. $\frac{k-1}{2k},k\notin (\frac{1}{3},3)$

Answer 1

The correct option is C $\frac{2k}{k-1},k\notin (\frac{1}{3},3)$

Using $\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$

We get $\frac{\tan 3A}{\tan A}=k=\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}$

So, ${\tan }^{2}A=\frac{k-3}{1-3k}=\frac{{\sin }^{2}A}{1-{\sin }^{2}A}$

So, ${\sin }^{2}A=\frac{k-3}{4(k-1)}$

Now, $\frac{\sin 3A}{\sin A}=\frac{3\sin A-4{\sin }^{3}A}{\sin A}=3-4{\sin }^{3}A=3-\frac{k-3}{1-k}$

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

Now, ${\tan }^{2}A>0$

So, $\frac{k-3}{3k-1}>0$

So, $k<\frac{1}{3}$ or $k>3$