Question

Prove that, $2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]=2\left[ \cot \dfrac{\pi }{4}-\cot \dfrac{5\pi }{12} \right]$

Answer 1

Hint: In this type of question we have to use the concept of trigonometry. Here, as we have to prove $2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]=2\left[ \cot \dfrac{\pi }{4}-\cot \dfrac{5\pi }{12} \right]$, it is sufficient to prove that, $\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right)=\cot \dfrac{5\pi }{12}$. To prove this we consider $\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right)$ and simplify it further then by using the formula $\cot \left( 2\pi +\theta \right)=\cot \theta $ we can obtain the required result.

Complete step by step answer:
Now we have to prove that, $2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]=2\left[ \cot \dfrac{\pi }{4}-\cot \dfrac{5\pi }{12} \right]$
For this let us consider,
\[\Rightarrow L.H.S.=2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]\cdots \cdots \cdots \left( e{{q}^{n}}1 \right)\]
Now, instead of considering the entire expression we will consider only the second term of the expression and let us simplify it.
Consider,
\[\Rightarrow \cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right)\]
We will simplify this expression by performing cross multiplication and making the denominator same we get,
\[\Rightarrow \cot \left( \dfrac{3\pi }{12}+\dfrac{26\pi }{12} \right)=\cot \left( \dfrac{29\pi }{12} \right)\]
Let us try to express \[\left( \dfrac{29\pi }{12} \right)\] in the form of \[\left( 2\pi +\theta \right)\] so that we are able to use the formula, \[\cot \left( 2\pi +\theta \right)=\cot \theta \]
\[\Rightarrow \cot \left( \dfrac{29\pi }{12} \right)=\cot \left( \dfrac{24\pi +5\pi }{12} \right)\]
\[\begin{align}
  & \Rightarrow \cot \left( \dfrac{29\pi }{12} \right)=\cot \left( \dfrac{24\pi }{12}+\dfrac{5\pi }{12} \right) \\
 & \Rightarrow \cot \left( \dfrac{29\pi }{12} \right)=\cot \left( 2\pi +\dfrac{5\pi }{12} \right) \\
\end{align}\]
Now, as we know that, \[\cot \left( 2\pi +\theta \right)=\cot \theta \] we can write,
\[\Rightarrow \cot \left( 2\pi +\dfrac{5\pi }{12} \right)=\cot \dfrac{5\pi }{12}\]
And hence we can say that,
\[\Rightarrow \cot \left( \dfrac{3\pi }{12}+\dfrac{26\pi }{12} \right)=\cot \dfrac{5\pi }{12}\]
Thus by substituting in \[\left( e{{q}^{n}}1 \right)\] we get,
\[\Rightarrow L.H.S.=2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]=2\left[ \cot \dfrac{\pi }{4}-\cot \dfrac{5\pi }{12} \right]\]
Therefore we have proved that, \[2\left[ \cot \dfrac{\pi }{4}-\cot \left( \dfrac{\pi }{4}+\dfrac{13\pi }{6} \right) \right]=2\left[ \cot \dfrac{\pi }{4}-\cot \dfrac{5\pi }{12} \right]\]

Note: In this type of question students may make mistake when they use the formula of cot as they are familiar with the formulas of sin, cos and tan so that they also have to remember the formula \[\cot \left( 2\pi +\theta \right)=\cot \theta \]. Some students try to simplify both sides by substituting the values using a calculator or web which is wrong as we have to prove the statement by using properties of trigonometry and not by calculator. Also students have to take care when they try to express \[\left( \dfrac{29\pi }{12} \right)\] in the form of \[\left( 2\pi +\theta \right)\].