The smallest positive value of \(x\) in degrees satisfying the equation \(\tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan (x) \tan \left(x-50^{\circ}\right)\) is

\(\begin{aligned} & \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan (x) \tan \left(x-50^{\circ}\right) \\ & \Rightarrow \frac{\tan \left(x+100^{\circ}\right)}{\tan \left(x-50^{\circ}\right)}=\tan \left(x+50^{\circ}\right) \tan (x) \\ & \Rightarrow \frac{2 \sin \left(x+100^{\circ}\right) \cos \left(x-50^{\circ}\right)}{2 \cos \left(x+100^{\circ}\right) \sin \left(x-50^{\circ}\right)}\end{aligned}\)
\(=\frac{2 \sin \left(x+50^{\circ}\right) \sin x}{2 \cos \left(x+50^{\circ}\right) \cos x}\)
\(\Rightarrow \frac{\sin (2 x+50)+\sin 150^{\circ}}{\sin (2 x+50)-\sin 150^{\circ}}\)
\(=\frac{\cos \left(50^{\circ}\right)-\cos \left(2 x+50^{\circ}\right)}{\cos \left(2 x+50^{\circ}\right)+\cos 50^{\circ}}\)
By componendo-dividendo, we get
\(\begin{aligned}
& \frac{2 \sin \left(2 x+50^{\circ}\right)}{2 \sin \left(150^{\circ}\right)}=\frac{2 \cos 50^{\circ}}{-2 \cos \left(2 x+50^{\circ}\right)} \\
& \Rightarrow 2 \sin \left(2 x+50^{\circ}\right) \cos \left(2 x+50^{\circ}\right) \\
& \Rightarrow=-2 \sin \left(150^{\circ}\right) \cos \left(50^{\circ}\right) \\
& \Rightarrow \sin \left(4 x+100^{\circ}\right)=-\cos 50^{\circ} \\
& \Rightarrow \sin \left(4 x+100^{\circ}\right)=\sin \left(270^{\circ}-50^{\circ}\right) \\
& \Rightarrow 4 x+100^{\circ}=220^{\circ} \\
& \Rightarrow 4 x=120^{\circ} \\
& \Rightarrow x=30^{\circ}
\end{aligned}\)