Question

If $\frac{\tan 3A}{\tan A}=a,$ then $\frac{\sin 3A}{\sin A}=$

• A. $\frac{2a}{a-1}$
• B. $\frac{2a}{a+1}$
• C. $\frac{a}{a+1}$
• D. $\frac{a}{a-1}$

Answer 1

Answer: (A)

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

Given, $\frac{\tan 3A}{\tan A}=a$

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

Or, $\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}=a$

Or, $a-3a{\tan }^{2}A=3-{\tan }^{2}A$

Or, ${\tan }^{2}A=\frac{a-3}{3a-1}$

$\frac{\sin 3A}{\sin A}=\frac{3\sin A-4{\sin }^{3}A}{\sin A}$

$=3-4{\sin }^{2}A$

$=3-4\frac{{\tan }^{2}A}{{\sec }^{2}A}=3-4\frac{{\tan }^{2}A}{1+{\tan }^{2}A}$

$=3-4\left(\frac{\frac{a-3}{3a-1}}{1+\frac{a-3}{3a-1}}\right)=3-\frac{a-3}{a-1}$

$=\frac{2a}{a-1}$ Ans