Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \cos (2 A)=2 \cos ^2 A-1 \\
& \cos (3 A)=4 \cos ^3 A-3 \cos A \\
& \sin (2 A)=2 \sin A \cos A \\
& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\
&
\end{aligned}
$$\begin{aligned}
S_1 & =\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)=\cot(10^{\circ})-\cot \left(50^{\circ}\right)+\tan(20^{\circ}) \\
\\
\Rightarrow S_1 & =\frac{\cos \left(10^{\circ}\right)}{\sin \left(10^{\circ}\right)}-\frac{\cos \left(50^{\circ}\right)}{\sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{2 \sin \left(10^{\circ}\right) \sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{\cos \left(40^{\circ}\right)-\frac{1}{2}}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\ \\
\Rightarrow S_1 & =\frac{4 \sin 40^{\circ}}{4 \cos ^2\left(20^{\circ}\right)-3}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}=\frac{6 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-3 \sin \left(20^{\circ}\right)}{0.5}
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow S_1=\frac{3}{0.5}\left(2 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-\sin \left(20^{\circ}\right)\right) \\ \\
& \Rightarrow S_1=6\left(\sin \left(60^{\circ}\right)\right)=3 \sqrt{3}
\end{aligned}$$
Best Answer
Since $$\tan{x}+\tan\left(60^{\circ}+x\right)+\tan\left(120^{\circ}+x\right)=$$ $$=\tan{x}+\frac{\sqrt3+\tan{x}}{1-\sqrt3\tan{x}}+\frac{-\sqrt3+\tan{x}}{1+\sqrt3\tan{x}}=3\tan3x,$$ we obtain: $$\cot10^{\circ}+\cot70^{\circ}-\cot50^{\circ}=\tan80^{\circ}+\tan20^{\circ}+\tan140^{\circ}=3\tan(3\cdot20^{\circ})=3\sqrt3.$$