Find the smallest positive value of $x$ ( in degrees ) for which $$\tan ( x + 100°) = \tan ( x + 50° ) \tan x \tan ( x - 50°)$$ My answer is too long and I don't think that this question deserves so long answer I will post my answer as soon as I got time, but need your help for a short answer. 
Sorry , guys for bad handwriting but I think that you will be able to know my way of this solution
We substitue $\alpha=\frac{50}{100}\pi$ then we get the equation
$$\tan(x+2\alpha)=\tan(x+\alpha)\tan(x)\tan(x-\alpha)$$, this equation can be simplified to
$$\left( \tan \left( x \right) \right) ^{3} \left( \tan \left( \alpha \right) \right) ^{4}+2\, \left( \tan \left( x \right) \right) ^{4} \tan \left( \alpha \right) -4\, \left( \tan \left( x \right) \right) ^{2} \left( \tan \left( \alpha \right) \right) ^{3}-\tan \left( x \right) \left( \tan \left( \alpha \right) \right) ^{4}- \left( \tan \left( x \right) \right) ^{3}+\tan \left( x \right) +2\,\tan \left( \alpha \right) =0$$ Now Substitute $$\tan(x)=t$$ and you will get an equation of degree four in $t$.