Question

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

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

Answer 1

Answer: (B)

Given that
$\frac{\tan 3 A}{\tan A}=a$
$\Rightarrow \frac{3 \tan A-\tan ^{3} A}{\tan A\left(1-3 \tan ^{2} A\right)}=a$
$\Rightarrow 3-\tan ^{2} A=a-3 a \tan ^{2} A$
$\Rightarrow \tan ^{2} A(3 a-1)=a-3$
$\Rightarrow \tan A=\pm \sqrt{\frac{a-3}{3 a-1}}$
Now,

$\frac{\sin 3 A}{\sin A}=\frac{3 \sin A-4 \sin ^{3} A}{\sin A}$
$=3-4 \sin ^{2} A=3-4\left(\frac{a-3}{4(a-1)}\right)$
$=\frac{3 a-3-a+3}{(a-1)}=\frac{2 a}{(a-1)}$