If sin⁡xsin⁡y=12,cos⁡xcos⁡y=32,where x,y∈0,π2, then the value of tan (x+y) is equal to

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

Solution:

Given, $\frac{\sin x}{\sin y} = \frac{1}{2}, \frac{\cos x}{\cos y} = \frac{3}{2}$

$\Rightarrow \frac{\tan x}{\tan y} = \frac{1}{3}$

$\begin{aligned} \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan xtan y} = \frac{4 \tan x}{1 - 3 \tan^{2} x} \\ Also, \sin y = 2 \sin x, \cos y = \frac{2}{3} \cos x \end{aligned}$

$\begin{array}{l} \Rightarrow \sin^{2} y + \cos^{2} y = 4 \sin^{2} x + \frac{4 \cos^{2} x}{9} = 1 \\ \Rightarrow 36 \tan^{2} x + 4 = 9 \sec^{2} x = 9 (1 + \tan^{2} x) \\ \Rightarrow 27 \tan^{2} x = 5 \\ \Rightarrow \tan x = \frac{\sqrt{5}}{3 \sqrt{3}} \\ \Rightarrow \tan (x + y) = \frac{\frac{4 \sqrt{5}}{3 \sqrt{3}}}{1 - \frac{15}{27}} = \sqrt{15} \end{array}$

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

Solution:

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