Let x, y ∈ (0, (π)/(2)) be such that (x+y) = 2 (x-y) and 5 (x+y) =...

Given $\sin(x+y) = 2\sin(x-y)$

$\Rightarrow \dfrac{\sin(x+y)}{\sin(x-y)} = \dfrac{2}{1}$

Applying componendo and dividendo:

$\dfrac{\sin(x+y)+\sin(x-y)}{\sin(x+y)-\sin(x-y)} = \dfrac{2+1}{2-1}$

$\Rightarrow \dfrac{2\sin x \cos y}{2\cos x \sin y} = 3$

$\Rightarrow \dfrac{\tan x}{\tan y} = 3 \dots (1)$

Also given $5\cos(x+y) = 3\cos(x-y)$

$\Rightarrow \dfrac{\cos(x+y)}{\cos(x-y)} = \dfrac{3}{5}$

Applying componendo and dividendo:

$\dfrac{\cos(x+y)+\cos(x-y)}{\cos(x+y)-\cos(x-y)} = \dfrac{3+5}{3-5}$

$\Rightarrow \dfrac{2\cos x \cos y}{-2\sin x \sin y} = \dfrac{8}{-2}$

$\Rightarrow -\cot x \cot y = -4$

$\Rightarrow \tan x \tan y = \dfrac{1}{4} \dots (2)$

Multiplying equations (1) and (2):

\left(\dfrac{\tan x}{\tan y}\right) (\tan x \tan y) = 3 \times \dfrac{1}{4}$<br><br>$\Rightarrow \tan^2 x = \dfrac{3}{4}$<br><br>Answer:$\dfrac{3}{4}