Question

Simplify the given trigonometric expression: $\cot {{12}^{\circ }}\cot {{38}^{\circ }}\cot {{52}^{\circ }}\cot {{60}^{\circ }}\cot {{78}^{\circ }}$.

Answer 1

Hint: We first take the identity formula of trigonometry of $\tan \theta .\cot \theta =1$ and $\cot \left( {{90}^{\circ }}-\theta \right)=\tan \theta $. We use them in two different cases for $\theta ={{12}^{\circ }}$ and $\theta ={{38}^{\circ }}$. We find out the values of them and use those values in the given equation to find the solution of the problem.

Complete step-by-step solution:
We know the trigonometric form $\tan \theta =\dfrac{1}{\cot \theta }\Rightarrow \tan \theta .\cot \theta =1$
We also have $\cot \left( {{90}^{\circ }}-\theta \right)=\tan \theta $.
Now we place the value of ${{12}^{\circ }}$ and ${{38}^{\circ }}$ in place of $\theta $.
So, $\cot \left( {{90}^{\circ }}-{{12}^{\circ }} \right)=\cot {{78}^{\circ }}=\tan {{12}^{\circ }}$ and $\cot \left( {{90}^{\circ }}-{{38}^{\circ }} \right)=\cot {{52}^{\circ }}=\tan {{38}^{\circ }}$.
We have a direct value of $\cot \left( {{60}^{\circ }} \right)=\dfrac{1}{\sqrt{3}}$.
We place all these values in the given equation of $\cot {{12}^{\circ }}\cot {{38}^{\circ }}\cot {{52}^{\circ }}\cot {{60}^{\circ }}\cot {{78}^{\circ }}$.
$\begin{align}
  & \cot {{12}^{\circ }}\cot {{38}^{\circ }}\cot {{52}^{\circ }}\cot {{60}^{\circ }}\cot {{78}^{\circ }} \\
 & =\left( \cot {{12}^{\circ }}\cot {{78}^{\circ }} \right)\left( \cot {{52}^{\circ }}\cot {{38}^{\circ }} \right)\left( \cot {{60}^{\circ }} \right) \\
 & =\left( \cot {{12}^{\circ }}\tan {{12}^{\circ }} \right)\left( \tan {{38}^{\circ }}\cot {{38}^{\circ }} \right)\left( \dfrac{1}{\sqrt{3}} \right) \\
\end{align}$
Now we use the identity $\tan \theta .\cot \theta =1$ for $\theta ={{12}^{\circ }}$ and $\theta ={{38}^{\circ }}$.
So, $\tan \left( {{12}^{\circ }} \right).\cot \left( {{12}^{\circ }} \right)=1$ and $\tan \left( {{38}^{\circ }} \right).\cot \left( {{38}^{\circ }} \right)=1$
The equation becomes
$\begin{align}
  & \cot {{12}^{\circ }}\cot {{38}^{\circ }}\cot {{52}^{\circ }}\cot {{60}^{\circ }}\cot {{78}^{\circ }} \\
 & =\left( \cot {{12}^{\circ }}\tan {{12}^{\circ }} \right)\left( \tan {{38}^{\circ }}\cot {{38}^{\circ }} \right)\left( \dfrac{1}{\sqrt{3}} \right) \\
 & =\left( 1 \right)\left( 1 \right)\left( \dfrac{1}{\sqrt{3}} \right) \\
 & =\dfrac{1}{\sqrt{3}} \\
\end{align}$
Therefore, simplification value of $\cot {{12}^{\circ }}\cot {{38}^{\circ }}\cot {{52}^{\circ }}\cot {{60}^{\circ }}\cot {{78}^{\circ }}$ is $\dfrac{1}{\sqrt{3}}$.

Note: We need to remember not to change the usual values of ${{30}^{\circ }}$, ${{45}^{\circ }}$, ${{60}^{\circ }}$ as we have the exact value of those trigonometric angles. We use identity formulas for other non-trivial angles like ${{12}^{\circ }}$, ${{38}^{\circ }}$.