You should be able to solve this.

Tanning beds cause cancer. Solve that.

I'm going off on a tangent. In bed. In space. With mah kitteh.

I've played around with it for a bit and I'm not entirely sure if the trick is noticing that:
tan(50)*tan(60)*tan(70)=tan(50)+tan(60)+tan(70)
or:
tan(80)=cot(10), therefore we would want to show that
sin(10)*sin(50)*sin(60)*sin(70)=cos(10)*cos(50)*cos(60)*cos(70)
and then using product to sum and picking pairs of angles clevery to reach the equality, since both the product of sin and cos produces sums of cos.
The obvious choice is picking the same pairs on both sides which leads to:
sin(a)*sin(b)*sin(c)*sin(d)=cos(a)*cos(b)*cis(c)*cos(d) iff
cos(a-b)*cos(c+d)=-cos(a+b)*cos(c-d)

P30498
>tan(50)*tan(60)*tan(70)=tan(50)+tan(60)+tan(70)
Neat, didn't notice that myself. The identity came up while working on P28972.

P30499
Well, yeah. Triangle on that thread satisfies the conditions on PDF related.

Another try:
Write it as tan(60)=tan(80)*tan(40)*tan(20).
The rationalle is that now triple and double angle formulas could be used to express everything in terms of tan(20). Since tan(60) is known, tan(20) can be obtained using the triple angle formula, by solving the equation x^3-3*sqrt(3)*x^2-3*x+sqrt(3)=0.
After [spoiler:cheating] solution seems to be 2*sqrt(3)*sin(10)-2*cos(10)+sqrt(3) (the other 2 solutions are real but do not aren't in the (0,1) interval).
I haven't really tried pluging tan(20) into the right hand side because it seems like a nightmare.

P30721 fuck off, mongloids

*mongoloids

P30721
tan(40) can also be obtained with the triple angle formula since tan(120) is known, to get 4*cos(50)-sqrt(3), which seems friendlier to plug into the right hand side by using double and half angle formula.
[spoiler:I realize this is probably not the intended way to prove this].

P30724 How about sin40, sin40 and sin40? since the sum of the angles is 120

Well, actually the root thing helped.

As I mentioned above, we can use sqrt(3)=tan(60)=tan(240)=tan(420), and the triple angle formula to obtain tan(20), tan(80) and tan(140) since all of them are roots of x^3-3*sqrt(3)*x^2-3*x+sqrt(3)=0.

We don't really need the actual values since we only care about the product, and the last coefficient is the (negative) product of all roots, that is tan(20)*tan(80)*tan(140)=-sqrt(3)=-tan(60).

Using tan(x)=cot(90-x) and tan(-x)=-tan(x), we get
cot(70)*tan(80)*cot(-50)=-tan(60), which leads to the desired result.

P30881
Also, we would need to show that tan(20), tan(80) and tan(140) are all different, but it's easy to see that 0 < tan(20) < 1, 1 < tan(80) and -1 < tan(140) < 0.