Ratio of Trigonometric Functions tanx/tany = 1/3;  sin2x/sin2y = 3/4 where 0 < x,y< pi/2. What is the Value of 
tan2x/tan2y
I couldn't find the appropriate trig identities to use for this problem, and what relationship for sin and tan that we can exploit.
 A: HINT: use that $$\frac{\sin(2x)}{\sin(2y)}=\frac{\tan(x)(1+\tan(y)^2)}{(1+\tan(x)^2)\tan(y)}=\frac{3}{4}$$
and $$\tan(2x)=\frac{2\tan(x)}{1-\tan(x)^2}$$
$$\tan(2y)=\frac{2\tan(y)}{1-\tan(y)^2}$$
A: $$\frac{\tan x}{\tan y}=\frac{1}{3};\;\frac{\sin 2x}{\sin 2y}=\frac{3}{4}$$
set $\tan x =t;\;\tan y=u$
$$\sin 2x=\frac{2t}{1+t^2};\;\sin 2y=\frac{2u}{1+u^2}$$
$$\frac{t}{u}=\frac{1}{3}\to u=3t$$
$$\frac{2t}{1+t^2}\frac{1+u^2}{2u}=\frac{3}{4}$$
$$\frac{2t}{1+t^2}\frac{1+9t^2}{6t}=\frac{3}{4}$$
$$8t(1+9t^2)=18t(1+t^2)\to 8t(1+9t^2)-18t(1+t^2)=0\to 2t(4+36t^2-9-9t^2)=0$$
$$\;t=\pm\sqrt{\frac{5}{27}};\;u=\pm 3\sqrt{\frac{5}{27}}=\pm\sqrt{\frac{5}{3}}$$
it $t=0$ then $u=0$ which is impossible since $u$ is the denominator of the fraction
if $t=\pm\sqrt{\frac{5}{27}};\;x=\pm\arctan\sqrt{\frac{5}{27}}+k\pi,\;\forall k\in\mathbb{Z}$
$u=\pm\sqrt{\frac{5}{3}};\;y=\pm\arctan \sqrt{\frac{5}{3}}+h\pi ,\;\forall h\in\mathbb{Z}$
A: Let $\dfrac{\sin x}{\sin y}=\dfrac{\cos x}{3\cos y}=k\  \  \ \ (1)$
$$\dfrac34=\dfrac{2\sin x\cos x}{2\sin y\cos y}=\dfrac{2k\sin y\cdot3k\cos y}{2\sin y\cos y}\implies k^2=\dfrac14\  \ \ \ (2)$$
$(1)\implies k^2\{\sin^2y+(3\cos y)^2\}=1\iff k^2=\dfrac1{1+8\cos^2y}$
By $(1),(2):$  $$\cos^2y=\dfrac38\implies \tan y=+\sqrt{\dfrac83-1}$$ as $0<y<\dfrac\pi2$
Now we have $\tan x=\dfrac{\tan y}3$
Finally use Double-Angle Formula of $\tan$
