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

We can take each of the terms in the LHS and simplify them. Then we can write the LHS in the simplified terms , using $\cos \left( {\pi + x} \right) = - \cos x$,$\cos \left( { - x} \right) = \cos x$,$\sin \left( {\pi - x} \right) = \sin x$,$\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x$,$\cot x = \dfrac{{\cos x}}{{\sin x}}$, and then simplify to get the required RHS. As the LHS and RHS are equal, we can say that the equation is true.

Complete step-by-step answer:
We need to prove that $\dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} = {\cot ^2}x$
So, we can take the LHS,
$LHS = \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}}$
We know that,
$\cos \left( {\pi + x} \right) = - \cos x$
$\cos \left( { - x} \right) = \cos x$
$\sin \left( {\pi - x} \right) = \sin x$
$\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x$
We can substitute the above equation in the LHS
$\Rightarrow LHS = \dfrac{{ - \cos \left( x \right)\cos \left( x \right)}}{{\sin \left( x \right)\left( { - \sin \left( x \right)} \right)}}$
On simplifying, we get,
$LHS = \dfrac{{{{\cos }^2}\left( x \right)}}{{{{\sin }^2}\left( x \right)}}$
Using the relation $\cot x = \dfrac{{\cos x}}{{\sin x}}$, we get,
$LHS = {\cot ^2}x$
As the RHS is also ${\cot ^2}x$ we can write,
$LHS = RHS$.
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
1. $\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x$
2.$\cos \left( {\pi + x} \right) = - \cos x$
3.$\cos \left( { - x} \right) = \cos x$
4.$\sin \left( {\pi - x} \right) = \sin x$
Adding $\pi$ or multiples of $\pi$ with the angle retains the ratio and adding $\dfrac{\pi }{2}$ or odd multiples of $\dfrac{\pi }{2}$ will change the ratio. While converting the angles we must take care of the sign of the ratio in its respective quadrant. In the 1st quadrant all the trigonometric ratios are positive. In the 2nd quadrant only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The following figure gives us an idea about the signs of different trigonometric functions. The angle measured in the counter clockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.