Question

If $\frac{\sin x}{\sin y}=\frac{1}{2},\frac{\cos x}{\cos y}=\frac{3}{2}$ where $x,y\in (0,\frac{\pi }{2})$ and the value of $\tan (x+y)$ is $k$, then $\left[k\right]=$
(where [.] denotes greatest integer function)

Answer 1

$\frac{\sin x}{\sin y}=\frac{1}{2},\frac{\cos x}{\cos y}=\frac{3}{2}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{1}{3}...\left(1\right)$

Now,
$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$
From equation $\left(1\right)$, we get
$=\frac{4\tan x}{1-3{\tan }^{2}x}$

Also,
$\sin y=2\sin x,\cos y=\frac{2}{3}\cos x$
$\Rightarrow {\sin }^{2}y+{\cos }^{2}y=1\Rightarrow 4{\sin }^{2}x+\frac{4}{9}{\cos }^{2}x=1\Rightarrow 36{\sin }^{2}x+4{\cos }^{2}x=9\Rightarrow 36{\tan }^{2}x+4=9{\sec }^{2}x\Rightarrow 36{\tan }^{2}x+4=9(1+{\tan }^{2}x)\Rightarrow 27{\tan }^{2}x=5\Rightarrow \tan x=\frac{\sqrt{5}}{3\sqrt{3}}(\because x\in (0,\pi /2\left)\right)\therefore \tan (x+y)=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-\frac{15}{27}}=\sqrt{15}\Rightarrow k=\sqrt{15}\therefore \left[k\right]=3$