Question

If $\frac{sinx}{siny}=\frac{1}{2},\frac{cosx}{cosy}=\frac{3}{2},x,yϵ(0,\frac{\pi }{2})$, then tan (x+y)=___

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

Answer 1

The correct option is D

$\sqrt{15}$

$\frac{sinx}{siny}=\frac{1}{2},\frac{cosx}{cosy}=\frac{3}{2}2sinx=siny,2cosx=3cosy\therefore \frac{tanx}{tany}=\frac{1}{3}\Rightarrow tany=3tanx.$
Now, $tan(x+y)=\frac{tanx+tany}{1-tanxtany}.$
$=\frac{tanx+3tanx}{1-tanx\left(3tanx\right)}=\frac{4tanx}{1-3ta{n}^{2}x}....\left(1\right)$
Also, sin y =2 sin x, cosy =$\frac{2}{3}cosx.$
$si{n}^{2}y+co{s}^{2}y=1\Rightarrow 4si{n}^{2}x+\frac{4}{9}co{s}^{2}x=1\Rightarrow 36ta{n}^{2}x+4=9se{c}^{2}x\Rightarrow 36ta{n}^{2}x+4=9(1+ta{n}^{2}x)\therefore 27ta{n}^{2}x=5\Rightarrow tanx=\frac{\sqrt{5}}{3\sqrt{3}}$
$\Rightarrow tan(x+y)=\frac{\frac{4\sqrt{5}}{3\sqrt{3}}}{1-\frac{15}{27}}=\frac{4\sqrt{5}.27}{12.3\sqrt{3}}=\sqrt{5}\sqrt{3}=\sqrt{15}$