Question

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

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

Answer 1

The correct option is D $\sqrt{15}$

As $\frac{\sin x}{\sin y}=\frac{1}{2}$ and $\frac{\cos x}{\cos y}=\frac{3}{2}$, then,

$\frac{\tan x}{\tan y}=\frac{\frac{\sin x}{\cos x}}{\frac{\sin y}{\cos y}}$

$=\frac{\sin x}{\sin y}\times \frac{\cos y}{\cos x}$

$=\frac{1}{2}\times \frac{2}{3}$

$=\frac{1}{3}$

$\tan y=3\tan x$

$\tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

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

$\tan \left(x+y\right)=\frac{4\tan x}{1-3{\tan }^{2}x}$ (1)

Since, $\frac{\sin x}{\sin y}=\frac{1}{2}$ and $\frac{\cos x}{\cos y}=\frac{3}{2}$, then,

$\sin y=2\sin x$ and $\cos y=\frac{2}{3}\cos x$

${\sin }^{2}y=4{\sin }^{2}x$ and ${\cos }^{2}y=\frac{4}{9}{\cos }^{2}x$

Applying the trigonometric identity,

${\sin }^{2}y+{\cos }^{2}y=1$

$4{\sin }^{2}x+\frac{4}{9}{\cos }^{2}x=1$

$4{\cos }^{2}x\left(\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{1}{9}\right)=1$

$\frac{9{\tan }^{2}x+1}{9}=\frac{1}{4{\cos }^{2}x}$

$\frac{9{\tan }^{2}x+1}{9}=\frac{{\sec }^{2}x}{4}$

$\frac{9{\tan }^{2}x+1}{9}=\frac{1+{\tan }^{2}x}{4}$

$36{\tan }^{2}x+4=9+9{\tan }^{2}x$

$27{\tan }^{2}x-5=0$

${\tan }^{2}x=\frac{5}{27}$

$\tan x=\frac{\sqrt{5}}{3\sqrt{3}}$

Put the value of $\tan x$ in equation (1),

$\tan \left(x+y\right)=\frac{4\left(\frac{\sqrt{5}}{3\sqrt{3}}\right)}{1-3{\left(\frac{\sqrt{5}}{3\sqrt{3}}\right)}^2}$

$=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-3\left(\frac{5}{27}\right)}$

$=\frac{3\sqrt{5}}{\sqrt{3}}$

$=\sqrt{15}$