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.   
 A: 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 $x=55^0$
A: Another way.
We need to solve $$\tan(x+100^{\circ})\cot{x}=\tan(x-50^{\circ})\tan(x+50^{\circ})$$ or
 $$\tan(x+100^{\circ})\cot{x}-1=\tan(x-50^{\circ})\tan(x+50^{\circ})-1$$ or
$$\frac{\sin100^{\circ}}{\cos(100^{\circ}+x)\sin{x}}+\frac{\cos2x}{\cos(x-50^{\circ})\cos(x+50^{\circ})}=0$$ or
$$\sin100^{\circ}(\cos2x+\cos100^{\circ})+\cos2x(\sin(2x+100^{\circ})-\sin100^{\circ})=0$$ or
$$\sin200^{\circ}+\sin(100^{\circ}+4x)+\sin100^{\circ}=0$$ or
$$2\sin150^{\circ}\cos50^{\circ}+\sin(100^{\circ}+4x)=0$$ or
$$\sin(100^{\circ}+4x)=\sin220^{\circ},$$
which gives
$$4x=120^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$, which is
$$x=30^{\circ}+90^{\circ}k$$ or
$$4x=-140^{\circ}+360^{\circ}k,$$ which is
$$x=-35^{\circ}+90^{\circ}k$$ and we got the answer: $30^{\circ}.$
A: I'd like to solve the same sum without skipping steps as done by the previous solver.
$\tan(x+100)=\tan(x+50)\tan x\tan(x-50)$
$\tan(x+100)/\tan(x-50) = \tan(x+50)\tan x$
$\sin(x+100)\cos(x-50)/\cos(x+100)\sin(x-50)=\sin(x+50)\sin x/\cos(x+50)\cos x$
Componendo and Dividendo,
$\sin(x+100)\cos(x-50)+\cos(x+100)\sin(x-50)/
\sin(x+100)\cos(x-50)-\cos(x+100)\sin(x-50)=\sin(x+50)\sin x+\cos(x+50)\cos x/\sin(x+50)\sin x-\cos(x+50)\cos x$
$\sin(x+100+x-50)/\sin(x+100-x+50)=\cos(x+50-x)/-\cos(x+50+x)$
$\sin(2x+50)/\sin 150 = \cos(2x+50)/-$cos50$
$\sin(2x+50)/1/2=-\cos 50/\cos(2x+50) (\sin150 = \sin(180-30) = \sin 30 =1/2)$
$2\sin(2x+50)\cos(2x+50) = -\cos 50$
$\sin(4x+100) = -\sin 40 (\sin 2x = 2\sin x\cos x)$
$\sin(4x+100) = \sin(180+40)$ (3rd quadrant, $\sin x$ is negative)
$4x+100 = 220$
$x=120/4 = 30$
Hope that made it easier :)
