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 $\tan (x+y)=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-\frac{15}{27}}=\frac{4\sqrt{5}\times 27}{12\times 3\sqrt{3}}$
$=\sqrt{15}$