Dig in trigo

Consider the following identity

$\tan(\alpha\pm\beta)=\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$

So we have

$\begin{aligned} \tan(2x)&=\tan[(x+y)+(x-y)]\\ &=\dfrac{\mathbin{\color{#D61F06}\tan(x+y)}+\mathbin{\color{#3D99F6}\tan(x-y)}}{1-\mathbin{\color{#D61F06}\tan(x+y)}\mathbin{\color{#3D99F6}\tan(x-y)}}\\ &=\dfrac{\mathbin{\color{#D61F06} 2\sqrt{2}}+\mathbin{\color{#3D99F6} \sqrt{2}}}{1-\mathbin{\color{#D61F06} 2\sqrt{2}}\cdot\mathbin{\color{#3D99F6} \sqrt{2}}}=-\dfrac{3\sqrt{2}}{3}=-\sqrt{2} \end{aligned}$

$\begin{aligned} \tan(2y)&=\tan[(x+y)-(x-y)]\\ &=\dfrac{\mathbin{\color{#D61F06}\tan(x+y)}-\mathbin{\color{#3D99F6}\tan(x-y)}}{1+\mathbin{\color{#D61F06}\tan(x+y)}\mathbin{\color{#3D99F6}\tan(x-y)}}\\ &=\dfrac{\mathbin{\color{#D61F06} 2\sqrt{2}}-\mathbin{\color{#3D99F6} \sqrt{2}}}{1+\mathbin{\color{#D61F06} 2\sqrt{2}}\cdot\mathbin{\color{#3D99F6} \sqrt{2}}}=\dfrac{\sqrt{2}}{5} \end{aligned}$

Our answer is $-\sqrt{2} \times \dfrac{\sqrt{2}}{5} = - \dfrac25 = \boxed{-0.4}$ .

$\begin{cases} \tan (x+y) = \sqrt 8 & \implies \sin (x+y) = \frac {\sqrt 8}3, & \cos (x+y) = \frac 13 \\ \tan (x-y) = \sqrt 2 & \implies \sin (x-y) = \sqrt {\frac 23}, & \cos (x-y) = \frac 1{\sqrt 3} \end{cases}$

$\begin{aligned} \sin(x+y) \sin(x-y) & = \frac {\sqrt 8}3 \cdot \sqrt {\frac 23} & \small \color{#3D99F6} \text{Since }\sin A \sin B = \frac 12 (\cos (A-B)-\cos (A+B)) \\ \frac 12 \left(\cos 2y - \cos 2x \right) & = \frac 4{3\sqrt 3} \\ \cos 2y - \cos 2x & = \frac 8{3\sqrt 3} & ...(1) \\ \cos(x+y) \cos(x-y) & = \frac 13 \cdot \frac 1{\sqrt 3} & \small \color{#3D99F6} \text{Since }\cos A \cos B = \frac 12 (\cos (A-B) + \cos (A+B)) \\ \frac 12 \left(\cos 2y + \cos 2x \right) & = \frac 1{3\sqrt 3} \\ \cos 2y + \cos 2x & = \frac 2{3\sqrt 3} & ...(2) \end{aligned}$

$\begin{cases} (2)+(1): \quad \cos 2y = \frac 5{\sqrt{27}} & \implies \sin 2y = \sqrt {\frac 2{27}}, & \color{#3D99F6} \tan 2y = \frac {\sqrt 2}5 \\ (2)-(1): \quad \cos 2x = - \frac 1{\sqrt 3} & \implies \sin 2x = \sqrt {\frac 23}, & \color{#D61F06} \tan 2x = - \sqrt 2 \end{cases}$

Then we have:

$\begin{aligned} \ {\color{#D61F06}\tan 2x} \ {\color{#3D99F6} \tan 2y} & = \boxed{-0.4} \end{aligned}$