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Prove the following-
$\cot 3{\text{A}} = \dfrac{{3cotA - {{\cot }^3}{\text{A}}}}{{1 - 3{{\cot }^2}{\text{A}}}}$


Answer
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Hint: We need to convert the multiples angle into half angles in this question. As we can see, cot3A is being written in terms of cotA, this means that we need to apply the angle sum formula for the cotangent function, which is given as-
$\cot \left( {{\text{A}} + {\text{B}}} \right) = \dfrac{{cotAcotB - 1}}{{cotA + cotB}}...\left( 1 \right)$

Complete step-by-step answer:
We need to find the value of cot3A.
We know that-
$\begin{align}
  &3{\text{A}} = 2{\text{A}} + {\text{A}} \\
  &\cot 3{\text{A}} = \cot \left( {2{\text{A}} + {\text{A}}} \right) \\
 & Applying\;property\;\left( 1 \right)\;we\;get, \\
 & \cot 3{\text{A}} = \dfrac{{\cot 2AcotA - 1}}{{\cot 2{\text{A}} + cotA}} \\
\end{align} $

Now we need to further simplify this, so we will convert cot2A into cotA as-
$\begin{align}
  &\cot 2{\text{A}} = \cot \left( {{\text{A}} + {\text{A}}} \right) \\
  &\cot 2{\text{A}} = \dfrac{{{{\cot }^2}{\text{A}} - 1}}{{2cotA}}...\left( 2 \right) \\
\end{align} $


So, we will use the equation (2) in the value of cot3A as-
$\begin{align}
  &\cot 3{\text{A}} = \dfrac{{\left( {\dfrac{{{{\cot }^2}{\text{A}} - 1}}{{2cotA}}} \right)cotA - 1}}{{\left( {\dfrac{{{{\cot }^2}{\text{A}} - 1}}{{2cotA}}} \right) + cotA}} \\
  &\cot 3{\text{A}} = \dfrac{{\dfrac{{{{\cot }^2}{\text{A}} - 1 - 2}}{2}}}{{\dfrac{{{{\cot }^2}{\text{A}} - 1 + 2{{\cot }^2}{\text{A}}}}{{2cotA}}}} = \dfrac{{cotA\left( {{{\cot }^2}{\text{A}} - 3} \right)}}{{3{{\cot }^2}{\text{A}} - 1}} \\
  &\cot 3{\text{A}} = \dfrac{{{{\cot }^3}{\text{A}} - 3cotA}}{{3{{\cot }^2}{\text{A}} - 1}} = \dfrac{{3cotA - {{\cot }^3}{\text{A}}}}{{1 - 3{{\cot }^2}{\text{A}}}} \\
\end{align} $

Hence, proved.

Note: In such types of questions, we need to analyze what type of conversions we need to make. For example, in this question we needed to convert cot3A into cotA, so we used angle sum property. Also, one should remember these properties for all trigonometric functions.