Question

Prove the following:
$\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}={{\cot }^{2}}x$

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like $\cos \left( \pi +\theta \right)=-\cos \theta $ , $\sin \left( \pi -\theta \right)=\sin \theta $ and $\cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin \theta $ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:
Given:
We have to prove the following equation:
$\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}={{\cot }^{2}}x$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \cos \left( \pi +\theta \right)=-\cos \theta ............\left( 1 \right) \\
 & \cos \left( -\theta \right)=\cos \theta ..................\left( 2 \right) \\
 & \sin \left( \pi -\theta \right)=\sin \theta ...............\left( 3 \right) \\
 & \cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin \theta ...........\left( 4 \right) \\
 & \dfrac{\cos \theta }{\sin \theta }=\cot \theta .......................\left( 5 \right) \\
\end{align}$
Now, we will use the above five formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}$ .
Now, we will use the formula from the equation (1) to write $\cos \left( \pi +x \right)=-\cos x$ and formula from the equation (2) to write $\cos \left( -x \right)=\cos x$ in the term on the left-hand side. Then,
$\begin{align}
  & \dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} \\
 & \Rightarrow \dfrac{-\cos x\times \cos x}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} \\
 & \Rightarrow \dfrac{-{{\cos }^{2}}x}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} \\
\end{align}$
Now, we will use the formula from the equation (3) to write $\sin \left( \pi -x \right)=\sin x$ and formula from the equation (4) to write $\cos \left( \dfrac{\pi }{2}+x \right)=-\sin x$ in the above expression. Then,
$\begin{align}
  & \dfrac{-{{\cos }^{2}}x}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} \\
 & \Rightarrow \dfrac{-{{\cos }^{2}}x}{\sin x\times \left( -\sin x \right)} \\
 & \Rightarrow \dfrac{-{{\cos }^{2}}x}{-{{\sin }^{2}}x} \\
 & \Rightarrow {{\left( \dfrac{\cos x}{\sin x} \right)}^{2}} \\
\end{align}$
Now, we will use the formula from the equation (5) to write $\dfrac{\cos x}{\sin x}=\cot x$ in the above expression. Then,
$\begin{align}
  & {{\left( \dfrac{\cos x}{\sin x} \right)}^{2}} \\
 & \Rightarrow {{\left( \cot x \right)}^{2}} \\
 & \Rightarrow {{\cot }^{2}}x \\
\end{align}$
Now, from the above result, we conclude that the value of the expression $\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}$ will be equal to the value of the expression ${{\cot }^{2}}x$ . Then,
$\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}={{\cot }^{2}}x$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)}={{\cot }^{2}}x$ .
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result. After that, we should apply trigonometric formulas like $\cos \left( \pi +\theta \right)=-\cos \theta $ and $\sin \left( \pi -\theta \right)=\sin \theta $ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.