Question

If \[x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }\], then prove that \[\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta }\] is also equal to x.

Answer 1

Hint: In this question, we have to somehow prove that \[\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }=\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta }\]. So, here we will use a few trigonometric identities to prove the desired relation such as \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Also, we need to know the concept of rationalization. After this information, we will be able to prove the required result.

Complete step-by-step answer:

In this question, we are asked to prove that \[\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta }\] is equal to x if \[x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }\]. So, in this question, we have to prove that \[\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }=\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta }\]. To prove the desired result, we will first consider the left-hand side of the relation. So, we can write it as

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

Now, we will rationalize the denominator by multiplying numerator and denominator by \[1+\sin \theta -\cos \theta \]. So, we get LHS as

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

Now, we know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. So, we can write LHS as

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

Now, we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. By using this, we can write the LHS as

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

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

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So we can also say that \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]. By using this relation, we will get LHS as

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

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

Now, we can see that \[2\sin \theta \] is common in both the terms of the denominator, so we can take it out as common. Therefore, we get LHS as

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

Now, we can see that \[2\sin \theta \] is common in both the numerator and denominator. So, we can cancel them. Therefore, we get LHS as

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

LHS = RHS

Hence, we have proved that

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


Note: In this question, one might ask why we did rationalization and how we will get to know that we have to rationalize. So, there is a trick, in the denominator of LHS, we were given \[\left( 1+\cos \theta +\sin \theta \right)\] and in the numerator, we were given \[\left( 1-\cos \theta +\sin \theta \right)\]. Then the only difference in both the terms was of the sign of \[\cos \theta \]. So, we wrote the term as \[\left[ \left( 1+\sin \theta \right)+\cos \theta \right]\] in the denominator and we rationalized it.