Question

Prove that $\cot A+\cot \left( {{60}^{\circ }}+A \right)+\cos \left( {{120}^{\circ }}+A \right)=3\cot 3A$.

Answer 1

In this question, we have to prove $\cot A+\cot \left( {{60}^{\circ }}+A \right)+\cos \left( {{120}^{\circ }}+A \right)=3\cot 3A$.
For solving this, we will use the various trigonometric formula given by:
(i) Summation of angles in cot function is given as $\cot \left( A+B \right)=\dfrac{\cot A\cdot \cot B-1}{\cot B+\cot A}$.
(ii) $\cot {{60}^{\circ }}$ is equal to $\dfrac{1}{\sqrt{3}}$.
(iii) $\cot {{120}^{\circ }}=\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)$. $\cot \theta$ is negative in II quadrant so $\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)=-\cot {{60}^{\circ }}=\dfrac{-1}{\sqrt{3}}$.
(iv) $\cot 3\theta =\dfrac{{{\cot }^{3}}\theta -3\cot \theta }{3{{\cot }^{2}}\theta -1}$.

Complete step-by-step solution
Here we are given the equation as $\cot A+\cot \left( {{60}^{\circ }}+A \right)+\cos \left( {{120}^{\circ }}+A \right)=3\cot 3A$.
We need to prove the left-hand side to be equal to the right-hand side.
Taking left hand side, $\cot A+\cot \left( {{60}^{\circ }}+A \right)+\cos \left( {{120}^{\circ }}+A \right)$.
We know that $\cot \left( A+B \right)=\dfrac{\cot A\cdot \cot B-1}{\cot B+\cot A}$.
So let us apply this on $\cot \left( {{60}^{\circ }}+A \right)\text{ and }\cos \left( {{120}^{\circ }}+A \right)$ we get:
$\cot A+\dfrac{\cot {{60}^{\circ }}\cdot \cot A-1}{\cot A+\cot {{60}^{\circ }}}+\dfrac{\cot {{120}^{\circ }}\cdot \cot A-1}{\cot A+\cot {{120}^{\circ }}}$.
We know from trigonometric ratio table that $\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}$ and $\cot {{120}^{\circ }}$ can be written as $\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)$ so we get:
$\cot A+\dfrac{\dfrac{1}{\sqrt{3}}\cot A-1}{\cot A+\dfrac{1}{\sqrt{3}}}+\dfrac{\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)\cdot \cot A-1}{\cot A+\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)}$.
We know that, cotangent is negative in II quadrant and ${{180}^{\circ }}-\theta$ will bring cot in II quadrant, so $\cot \left( {{180}^{\circ }}-\theta \right)$ will be equal to $-\cot \theta$.
Hence, $\cot \left( {{180}^{\circ }}-{{60}^{\circ }} \right)=-\cot {{60}^{\circ }}=\dfrac{-1}{\sqrt{3}}$ putting in the value we get:
$\cot A+\dfrac{\dfrac{1}{\sqrt{3}}\cot A-1}{\cot A+\dfrac{1}{\sqrt{3}}}+\dfrac{-\dfrac{1}{\sqrt{3}}\cot A-1}{\cot A-\dfrac{1}{\sqrt{3}}}$.
Now let us take LCM in the denominator for second and third term, we get:
$\cot A+\dfrac{\left( \dfrac{1}{\sqrt{3}}\cot A-1 \right)\left( \cot A-\dfrac{1}{\sqrt{3}} \right)+\left( -\dfrac{1}{\sqrt{3}}\cot A-1 \right)\left( \cot A+\dfrac{1}{\sqrt{3}} \right)}{\left( \cot A+\dfrac{1}{\sqrt{3}} \right)\left( \cot A-\dfrac{1}{\sqrt{3}} \right)}$.
Applying $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the denominator on the second term and simplifying the numerator of the second term by opening bracket, we get:
$\cot A+\dfrac{\left( \dfrac{1}{\sqrt{3}}{{\cot }^{2}}A-\cot A-\dfrac{1}{3}\cot A+\dfrac{1}{\sqrt{3}} \right)+\left( -\dfrac{1}{\sqrt{3}}{{\cot }^{2}}A-\cot A-\dfrac{1}{3}\cot A-\dfrac{1}{\sqrt{3}} \right)}{{{\cot }^{2}}A-\dfrac{1}{3}}$.
Further simplifying the numerator of the second term we get:
$\cot A+\dfrac{\left( -2\cot A-\dfrac{2}{3}\cot A \right)}{{{\cot }^{2}}A-\dfrac{1}{3}}$.
Now taking LCM in the denominator of both terms we get:
$\begin{aligned} & \dfrac{\cot A\left( {{\cot }^{2}}A-\dfrac{1}{3} \right)-2\cot A-\dfrac{2}{3}\cot A}{{{\cot }^{2}}A-\dfrac{1}{3}} \\\ & \Rightarrow \dfrac{{{\cot }^{3}}A-\dfrac{1}{3}\cot A-2\cot A-\dfrac{2}{3}\cot A}{{{\cot }^{2}}A-\dfrac{1}{3}} \\\ \end{aligned}$.
Now adding $\dfrac{1}{3}\cot A\text{ and }\dfrac{2}{3}\cot A$ we get:
$\Rightarrow \dfrac{{{\cot }^{3}}A-\dfrac{3}{3}\cot A-2\cot A}{{{\cot }^{2}}A-\dfrac{1}{3}}$.
Cancelling 3 we get:
$\begin{aligned} & \Rightarrow \dfrac{{{\cot }^{3}}A-\cot A-2\cot A}{{{\cot }^{2}}A-\dfrac{1}{3}} \\\ & \Rightarrow \dfrac{{{\cot }^{3}}A-3\cot A}{{{\cot }^{2}}A-\dfrac{1}{3}} \\\ \end{aligned}$.
Taking LCM of 3 in the denominator of the denominator of the term we get:
$\Rightarrow \dfrac{{{\cot }^{3}}A-3\cot A}{\dfrac{3{{\cot }^{2}}A-1}{3}}$.
Taking 3 in the numerator we get:
$\Rightarrow 3\left( \dfrac{{{\cot }^{3}}A-3\cot A}{3{{\cot }^{2}}A-1} \right)$.
Now we know that $\cot 3\theta =\dfrac{{{\cot }^{3}}\theta -3\cot \theta }{3{{\cot }^{2}}\theta -1}$ so we get:
$3\cot A$.
Which is equal to the right-hand side.
Hence proved.

Note: This sum is very tricky with lots of complex calculations, so students should do it carefully step by step without shortcuts. There are huge chances of making mistakes in positive or negative signs. Take LCM carefully. Keep in mind all the trigonometric formulas for solving this sum. $\cot {{120}^{\circ }}$ can also be found using $\cot {{120}^{\circ }}=\cot \left( {{90}^{\circ }}+{{30}^{\circ }} \right)=-\tan {{30}^{\circ }}=\dfrac{-1}{\sqrt{3}}$.