Determine the smallest positive value of x(in degrees) for which: $\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$ Determine the smallest positive value of x(in degrees) for which:
$\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$
I tried to apply the formula of $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$ but that led me nowhere resulting in a huge equation.
Please help.   

Given $\displaystyle \tan(x+100^0) = \tan(x+50^0)\cdot \tan (x)\cdot \tan(x-50^0)$
$\displaystyle \Rightarrow \frac{\tan(x+100^0)}{\tan(x-50^0)}\Rightarrow  =\tan(x+50^0)\cdot \tan(x^0)$
$\displaystyle \Rightarrow \frac{\sin(x+100^0)\cdot\cos(x-50^0)}{\cos(x+100^0)\cdot \sin (x-50^0)}=\frac{\sin(x+50^0)\cdot\sin(x)}{\cos(x+50^0)\cdot \cos (x)}$
Using Componendo and dividendo,
$\displaystyle\Rightarrow \frac{\sin(2x+50^0)}{\sin (150^0)} = -\frac{\cos(50^0)}{\cos(2x+50^0)}$
$\displaystyle \Rightarrow \sin(4x+100^0)=-\cos(50^0) = -\sin (50^0)=\sin(180^0+40^0)=\sin(360^0-40^0)$
$\displaystyle \Rightarrow (4x+100^0)=220^0 = 320^0$
So $x=30^0$ or $x=55^0$
