Question

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

Answer 1

Hint: To prove this relation, if we start from the left-hand side of the expression, then we should know a few trigonometric relations like \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \]. By using these relations, we can prove the expression.

Complete step-by-step answer:

In this question, we have to prove that

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

To prove this relation, we will first consider the left-hand side of the desired result, so we can write it as,

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

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

Now, we can see that \[\cos \theta \] is common in both the terms of the numerator, so we can take it out as common. Therefore, we get,

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

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, we can also write it as \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]. Therefore, we will get

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

\[LHS=\dfrac{\left( 2\sin \theta -1 \right)\cos \theta }{2{{\sin }^{2}}\theta -\sin \theta }\]
Now, we can see that \[\sin \theta \] is common in both the terms of the denominator. So, after taking it out as common, we will get,

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

Now, we can see that \[\left( 2\sin \theta -1 \right)\] is common in both the numerator and denominator. So, we can cancel them out and we will get LHS as

\[LHS=\dfrac{\cos \theta }{\sin \theta }\]

Now, we know that \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \]. So, we will get,
\[LHS=\cot \theta \]

LHS = RHS

Hence proved


Note: In this question, one might think to prove the result from RHS. So, if we start from RHS, then we will apply the identity \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] and then we will multiply numerator and denominator by \[\left( 2\sin \theta -1 \right)\] , and after expansion, we will write \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]. And, this is how we will derive the result from RHS to LHS.