Question

If $\sin x+\sin y=3/4$ and $\sin x-\sin y=2/5$, then $\frac{\tan \left[\right(x-y\left)/2\right]}{\tan \left[\right(x+y\left)/2\right]}=?$

• A. $15/8$
• B. $8/15$
• C. $3/10$
• D. $10/3$

Answer 1

Answer: (B)

$8/15$

Solution:

$\sin x+\sin y=\frac{3}{4}-\left(1\right)$

$\sin x-\sin y=\frac{2}{5}-\left(2\right)$

$\frac{\tan \left(\frac{x-y}{2}\right)}{\tan \left(\frac{x+y}{2}\right)}=\frac{2\sin \left(\frac{x-y}{2}\right)\cos \left(\frac{x+y}{2}\right)}{2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}$

$\Rightarrow \frac{\tan \left(\frac{x-y}{2}\right)}{\tan \left(\frac{x+y}{2}\right)}=\frac{\sin \left(\frac{2x}{2}\right)+\sin \left(\frac{-2y}{2}\right)}{\sin \left(\frac{2x}{2}\right)+\sin \left(\frac{2y}{2}\right)}$

$\Rightarrow \frac{\tan \left(\frac{x-y}{2}\right)}{\tan \left(\frac{x+y}{2}\right)}=\frac{\sin x-\sin y}{\sin x+\sin y}=\frac{2}{5}\times \frac{4}{3}=\frac{8}{15}$

$\therefore \text{Answer: option (B)}$