<s>[INST] 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.   
 [/INST] 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 :)
</s>