Question

Prove the following trigonometric function:
$\cos \left( \dfrac{3\pi }{2}+\theta \right)\cos \left( 2\pi +\theta \right)\left[ \cot \left( \dfrac{3\pi }{2}-\theta \right)+\cot \left( 2\pi +\theta \right) \right]=1$

Answer 1

Hint:In this question, the angles of which the trigonometric ratios are to be found are given as a sum of an angle $\theta $ with a multiple of $\dfrac{\pi }{2}$. Therefore, we should use the relations of the trigonometric ratios of angles and those of the sum of the angles with a multiple of $\dfrac{\pi }{2}$ to solve the given expression and find the answer.

Complete step-by-step answer:
In this case, the angle $\theta $ appears in a sum with a multiple of $\dfrac{\pi }{2}$. Therefore, we can use the following trigonometric identities to solve this question.
(a) Any trigonometric ratio of an angle and that of the sum of the angle with any multiple of $2\pi $ is the same…………………………(1.1)
(b) As cos function is an even function and the sin function is an odd function, for any angle $\theta $,
$\cos (-\theta )=\cos (\theta )$
$\sin (-\theta )=\sin (\theta )...................(1.2)$
(c) The sine and cos of angles follow the following identities
$\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin (\theta )$
$\cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin (\theta )$
$\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \left( \theta \right).............(1.3)$

Therefore, using equations (1.1), (1.2) and (1.3), we get
$\cos \left( \dfrac{3\pi }{2}+\theta \right)=\cos \left( 2\pi -\dfrac{\pi }{2}+\theta \right)=\cos \left( -\left( \dfrac{\pi }{2}-\theta \right) \right)=\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin (\theta )......(1.4)$
And
$\begin{align}
  & \cot \left( \dfrac{3\pi }{2}-\theta \right)=\cot \left( 2\pi -\dfrac{\pi }{2}-\theta \right)=\cot \left( -\left( \dfrac{\pi }{2}+\theta \right) \right)=\dfrac{\cos \left( -\left( \dfrac{\pi }{2}+\theta \right) \right)}{\sin \left( -\left( \dfrac{\pi }{2}+\theta \right) \right)} \\
 & =\dfrac{\cos \left( \dfrac{\pi }{2}+\theta \right)}{-\sin \left( \dfrac{\pi }{2}+\theta \right)}=\dfrac{-\sin \left( \theta \right)}{-\cos (\theta )}=\dfrac{\sin \left( \theta \right)}{\cos (\theta )}=\dfrac{1}{\cot (\theta )}={\tan (\theta )}=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}.........(1.5) \\
\end{align}$

${\cot (2\pi + \theta )}={\cot (\theta )}=\dfrac{\cos \left( \theta \right)}{\sin \left( \theta \right)}$ from $1.1$
Therefore, using (1.1), (1.4) and (1.5), we get

\[\begin{align}
  & \cos \left( \dfrac{3\pi }{2}+\theta \right)\cos \left( 2\pi +\theta \right)\left[ \cot \left( \dfrac{3\pi }{2}-\theta \right)+\cot \left( 2\pi +\theta \right) \right] \\
 & =\sin \left( \theta \right)\cos \left( \theta \right)\left[ \dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}+\dfrac{\cos \left( \theta \right)}{\sin \left( \theta \right)} \right]=\sin \left( \theta \right)\cos \left( \theta \right)\left[ \dfrac{{{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)}{\cos \left( \theta \right)\sin \left( \theta \right)} \right]..........(1.6) \\
\end{align}\]

Now, cancelling out the factor \[\cos \left( \theta \right)\sin \left( \theta \right)\] from the numerator and the denominator and using the fact that \[{{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)=1\], we obtain
$\cos \left( \dfrac{3\pi }{2}+\theta \right)\cos \left( 2\pi +\theta \right)\left[ \cot \left( \dfrac{3\pi }{2}-\theta \right)+\cot \left( 2\pi +\theta \right) \right]=1$
Which is the expression we wanted to prove.

Note: In equation (1.6), we should note that all the terms should be written in terms of sin and cosine so that we can cancel out the numerator and denominator. We could also have written everything in the form of tan and cot, but the steps would have been more complex and cumbersome.Students should remember important trigonometric identities and formulas and also about odd and even trigonometric functions for solving these types of questions.