Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Given that,

$\frac{\tan \left(3A\right)}{\tan \left(A\right)}=a$

Then, $\frac{\sin \left(3A\right)}{\sin \left(A\right)}$

Multiply and divide by $2\cos \left(A\right)$,

$\therefore \frac{2\cos \left(A\right)\sin \left(3A\right)}{2\cos \left(A\right)\sin \left(A\right)}$

$=\frac{2\cos \left(A\right)\sin \left(3A\right)}{\sin \left(2A\right)}\left[\because \sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)\right]$

$$\begin{gathered} =\frac{2\cos \left(A\right)\sin \left(3A\right)}{\sin \left(3A-A\right)} \\ =\frac{2\cos \left(A\right)\sin \left(3A\right)}{\left[\sin \left(3A\right)\cos \left(A\right)-\cos \left(3A\right)\sin \left(A\right)\right]} \end{gathered}$$

Step 2: Divide the numerator and denominator by $\cos \left(A\right)\cos \left(3A\right)$,

$\begin{array}{rcl}& \therefore & \frac{{\displaystyle \frac{2\cos \left(A\right)\sin \left(3A\right)}{\cos \left(A\right)\cos \left(3A\right)}}}{{\displaystyle \frac{\left[\sin \left(3A\right)\cos \left(A\right)-\cos \left(3A\right)\sin \left(A\right)\right]}{\cos \left(A\right)\cos \left(3A\right)}}}\\ & =& \frac{2\tan \left(3A\right)}{\left[\tan \left(3A\right)-\tan \left(A\right)\right]}\end{array}$

Step 3: Divide the numerator and denominator by $\tan \left(A\right)$,

$\begin{array}{rcl}& & \end{array}$$$\begin{gathered} =\frac{{\displaystyle \frac{2\tan \left(3A\right)}{\tan \left(A\right)}}}{{\displaystyle \frac{\left[\tan \left(3A\right)-\tan \left(A\right)\right]}{\tan \left(A\right)}}} \\ =\frac{2a}{\frac{\tan \left(3A\right)}{\tan \left(A\right)}-1} \\ =\frac{2a}{a-1} \end{gathered}$$

Hence, the correct option is (B).