Question

Prove the following expression
\[1-\dfrac{{{\sin }^{2}}\theta }{1+\cot \theta }-\dfrac{{{\cos }^{2}}\theta }{1+\tan \theta }=\sin \theta \cos \theta \]

Answer 1

Hint: To prove this relation, we should know a few trigonometrical relations such as \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]. Also, we should remember that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By using these relations we can prove the desired result.

Complete step-by-step answer:

In this question, we are asked to prove that

\[1-\dfrac{{{\sin }^{2}}\theta }{1+\cot \theta }-\dfrac{{{\cos }^{2}}\theta }{1+\tan \theta }=\sin \theta \cos \theta \]

To prove this relation, we will first consider the left-hand side of the given relation. So, we get,

\[LHS=1-\dfrac{{{\sin }^{2}}\theta }{1+\cot \theta }-\dfrac{{{\cos }^{2}}\theta }{1+\tan \theta }\]

Now, we know that \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So, we will get the LHS by using these relations as,

\[LHS=1-\dfrac{{{\sin }^{2}}\theta }{1+\dfrac{\cos \theta }{\sin \theta }}-\dfrac{{{\cos }^{2}}\theta }{1+\dfrac{\sin \theta }{\cos \theta }}\]

Now, we will take the LCM in the denominator of the terms. So, we get,

\[LHS=1-\dfrac{{{\sin }^{2}}\theta }{\dfrac{\sin \theta +\cos \theta }{\sin \theta }}-\dfrac{{{\cos }^{2}}\theta }{\dfrac{\cos \theta +\sin \theta }{\cos \theta }}\]

On further simplification, we will get,

\[LHS=1-\dfrac{\left( {{\sin }^{2}}\theta \right)\left( \sin \theta \right)}{\sin \theta +\cos \theta }-\dfrac{\left( {{\cos }^{2}}\theta \right)\left( \cos \theta \right)}{\sin \theta +\cos \theta }\]

\[LHS=1-\dfrac{{{\sin }^{3}}\theta }{\sin \theta +\cos \theta }-\dfrac{{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }\]

Now, we will take the LCM of all the terms in the LHS. So, we will get LHS as

\[LHS=\dfrac{\left( \sin \theta +\cos \theta \right)-{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\left( \sin \theta +\cos \theta \right)}\]

\[LHS=\dfrac{\sin \theta -{{\sin }^{3}}\theta +\cos \theta -{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }\]

Now, we will take \[\sin \theta \] common from \[\sin \theta \] and \[{{\sin }^{3}}\theta \] and similarly \[\cos \theta \] common from \[\cos \theta \] and \[{{\cos }^{3}}\theta \]. So, we get LHS as

\[LHS=\dfrac{\sin \theta \left( 1-{{\sin }^{2}}\theta \right)+\cos \theta \left( 1-{{\cos }^{2}}\theta \right)}{\sin \theta +\cos \theta }\]

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, we can say that \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]. So, we get LHS by using these relations as,

\[LHS=\dfrac{\sin \theta \left( {{\cos }^{2}}\theta \right)+\cos \theta \left( {{\sin }^{2}}\theta \right)}{\sin \theta +\cos \theta }\]

Here, we can see that \[\left( \sin \theta \right)\left( \cos \theta \right)\] is common in both the terms of the numerator. So, we can take it out as common. Therefore, we get,

\[LHS=\dfrac{\left( \sin \theta \right)\left( \cos \theta \right)\left[ \cos \theta +\sin \theta \right]}{\sin \theta +\cos \theta }\]

Here, we can see that \[\sin \theta +\cos \theta \] is common in both the numerator and denominator. So, we can cancel them out and by canceling them, we will get LHS as
\[LHS=\sin \theta \cos \theta \]

LHS = RHS

Hence we have proved that

\[1-\dfrac{{{\sin }^{2}}\theta }{1+\cot \theta }-\dfrac{{{\cos }^{2}}\theta }{1+\tan \theta }=\sin \theta \cos \theta \]


Note: In this question, the possible mistake one can make is by not converting tan and cot in terms of sin and cos and by directly taking the LCM. That is not a wrong way to solve, but that will complicate the solution. Also, later on, we have to convert them in terms of sin and cos which can lead to silly mistakes. So, it’s better to convert them at the beginning itself.