Question

If $\frac{\sin x}{\sin y}=\frac{1}{2}, \frac{\cos x}{\cos y}=\frac{3}{2}$, where $x, y \in\left(0, \frac{\pi}{2}\right)$, then the value of $\tan (x+ y)$ is equal to

• A. $\sqrt{13}$
• B. $\sqrt{14}$
• C. $\sqrt{17}$
• D. $\sqrt{15}$

Answer 1

Answer: (D)

$\frac{\sin x}{\sin y}=\frac{1}{2}, \frac{\cos x}{\cos y}=\frac{3}{2}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{1}{3}$
$\Rightarrow \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{4 \tan x}{1-3 \tan ^{2} x}$
Also $\sin y=2 \sin x, \cos y=\frac{2}{3} \cos x$
or $\sin ^{2} y+\cos ^{2} y=4 \sin ^{2} x+\frac{4 \cos ^{2} x}{9}=1$
or $36 \tan ^{2} x+4=9 \sec ^{2} x=9\left(1+\tan ^{2} x\right)$
or $27 \tan ^{2} x=5$
or $\tan x=\frac{\sqrt{5}}{3 \sqrt{3}}$
or $\tan (x+y)=\frac{\frac{4 \sqrt{5}}{3 \sqrt{3}}}{1-\frac{15}{27}}=\frac{4 \sqrt{5} \times 27}{12 \times 3 \sqrt{3}}$
$=\sqrt{15}$