Question

Prove that $\dfrac{\cos \theta \cos \left( 90-\theta \right)}{\cot \left( 90-\theta \right)}={{\cos }^{2}}\theta$

Answer 1

We are given two expressions involving trigonometric terms in each part and we need to prove that one is equal to the other. For this we use several trigonometric formulae and we try to simplify the equation as much as possible. We will start with the left side and we then perform some operations on it to turn it into the expression written on the right hand side.

Complete step-by-step solution:
We are given on the left hand side:
$\dfrac{\cos \theta \cos \left( 90-\theta \right)}{\cot \left( 90-\theta \right)}$
We first resolve the bracket involving $\cos \left( 90-\theta \right)$.
We know that:
$\cos \left( 90-\theta \right)=\sin \theta$
Putting it in the numerator we obtain:
$\dfrac{\cos \theta \sin \theta }{\cot \left( 90-\theta \right)}$
For denominator, we use the formula:
$\cot \left( 90-\theta \right)=\tan \theta$
Putting this in denominator we obtain:
$\dfrac{\cos \theta \sin \theta }{\tan \theta }$
Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Putting this we obtain:
$\dfrac{\cos \theta \sin \theta }{\dfrac{\sin \theta }{\cos \theta }}={{\cos }^{2}}\theta$=RHS
So, we get LHS=RHS. Hence, proved.

Note: Most of the time, while using the trigonometric formulae, we may plug in the wrong equations. So, the trigonometric formulae must be learned properly so that you are able to prove the quality of expressions without any mistakes. Moreover, always start with the side that is complex. As in this question, if you start with the right hand side then you will have no operation to perform and you will get stuck, so always start with the complex side and try to reduce it to the expression that doesn’t involve much terms.