Prove that: \[ \cot x+\cot\left(\frac{\pi}{3}+x\right) +\cot\left(\frac{2\pi}{3}+x\right) = 3\cot 3x \]
Solution
Let
\[
t=\cot x
\]
Using
\[
\cot(A+B)
=
\frac{\cot A\cot B-1}{\cot A+\cot B}
\]
and
\[
\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}
\]
\[
\cot\frac{2\pi}{3}=-\frac{1}{\sqrt{3}}
\]
we get
\[
\cot\left(\frac{\pi}{3}+x\right)
=
\frac{\frac{t}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+t}
=
\frac{t-\sqrt{3}}{1+\sqrt{3}t}
\]
\[
\cot\left(\frac{2\pi}{3}+x\right)
=
\frac{-\frac{t}{\sqrt{3}}-1}{-\frac{1}{\sqrt{3}}+t}
=
-\frac{t+\sqrt{3}}{\sqrt{3}t-1}
\]
Now,
\[
LHS
=
t+\frac{t-\sqrt{3}}{1+\sqrt{3}t}
-\frac{t+\sqrt{3}}{\sqrt{3}t-1}
\]
Take the common denominator:
\[
(1+\sqrt{3}t)(\sqrt{3}t-1)
=
3t^2-1
\]
Therefore,
\[
LHS
=
\frac{
t(3t^2-1)
+(t-\sqrt{3})(\sqrt{3}t-1)
-(t+\sqrt{3})(1+\sqrt{3}t)
}
{3t^2-1}
\]
Expand:
\[
(t-\sqrt{3})(\sqrt{3}t-1)
=
\sqrt{3}t^2-4t+\sqrt{3}
\]
\[
(t+\sqrt{3})(1+\sqrt{3}t)
=
3t+\sqrt{3}t^2+t+\sqrt{3}
\]
Substituting,
\[
LHS
=
\frac{
3t^3-t+\sqrt{3}t^2-4t+\sqrt{3}
-3t-\sqrt{3}t^2-t-\sqrt{3}
}
{3t^2-1}
\]
\[
=
\frac{3t^3-9t}{3t^2-1}
\]
\[
=
\frac{3(t^3-3t)}{3t^2-1}
\]
Using the identity
\[
\cot 3x
=
\frac{t^3-3t}{3t^2-1}
\]
Hence,
\[
LHS=3\cot 3x
\]
Hence proved.