Question

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

Answer 1

Hint: A very important formula of $\cot \theta $ is used here. There is also requirement of value of $\cot \left( {{{60}^ \circ }} \right)$ and $\cot \left( {{{120}^ \circ }} \right)$ to simplify this question and there is also requirement of value of $\cot \left( {3\theta } \right)$. This tells us the importance of formulas in trigonometry and also in this question.
Formula used: $1.\,\,\cot \left( {C + D} \right) = \dfrac{{\cot C\cot D - 1}}{{\cot C + \cot D}}$
$2.\,\,\cot \,3\theta = \dfrac{{\cot 3\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}}$


Complete step-by-step answer:
In the given question,
We know that
$\,\cot \left( {C + D} \right) = \dfrac{{\cot C\cot D - 1}}{{\cot C + \cot D}}$
We have,
$L.H.S = \,\cot A + \cot \left( {{{60}^ \circ } + A} \right) + \cot \left( {{{120}^ \circ } + A} \right)$
Now, using the above formula
$ \Rightarrow \cot A + \dfrac{{\cot {{60}^ \circ } \times \cot A - 1}}{{\cot A + \cot {{60}^ \circ }}} + \dfrac{{\cot {{120}^ \circ } \times \cot A - 1}}{{\cot A + \cot {{120}^ \circ }}}$
We know that, $\cot {60^ \circ } = \dfrac{1}{{\sqrt 3 }}\,\,and\,\,\cot {120^ \circ } = \dfrac{{ - 1}}{{\sqrt 3 }}$
$ \Rightarrow \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, taking L.C.M
$ \Rightarrow \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)}}{{{{\cot }^2}A - \dfrac{1}{3}}}$
$ \Rightarrow \cot A + \dfrac{{\dfrac{1}{{\sqrt 3 }}{{\cot }^2}A - \cot A - \dfrac{1}{3}\cot A + \dfrac{1}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }}{{\cot }^2}A - \cot A - \dfrac{1}{3}\cot A - \dfrac{1}{{\sqrt 3 }}}}{{{{\cot }^2}A - \dfrac{1}{3}}}$
Again, taking L.C.M
$ \Rightarrow \dfrac{{{{\cot }^3}A - \dfrac{1}{3}\cot A + \dfrac{1}{{\sqrt 3 }}{{\cot }^2}A - \cot A - \dfrac{1}{3}\cot A + \dfrac{1}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }}{{\cot }^2}A - \cot A - \dfrac{1}{3}\cot A - \dfrac{1}{{\sqrt 3 }}}}{{\dfrac{{3{{\cot }^2}A - 1}}{3}}}$
$ \Rightarrow \dfrac{{{{\cot }^3}A - \dfrac{3}{3}\cot A - 2\cot A}}{{\dfrac{{3{{\cot }^2}A - 1}}{3}}}$
$ \Rightarrow \dfrac{{3\left( {{{\cot }^3}A - 3\cot A} \right)}}{{\left( {3{{\cot }^2}A - 1} \right)}}$
Now, using the formula
$\left[ {\,\cot \,3\theta = \dfrac{{\cot 3\theta - 3\cot \theta }}{{3{{\cot }^2}\theta - 1}}} \right]$
On substituting the value, we have
$ \Rightarrow 3\cot 3\theta = R.H.S$
Therefore, $L.H.S = R.H.S$
Hence, proved.

Note: While doing the questions of trigonometry one can remember that the most important thing here is the formula. If you have a very good grip on formula then you can easily master this topic. Also knowing the conversion of one trigonometric identity into another is a very important aspect. Remember the values of some specific degrees of all trigonometric functions