Question

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

Answer 1

Hint: In this question, we will start from the right-hand side of the equation. We will rationalize the denominator by multiplying the numerator and denominator by \[\left( 1+\cos \theta +\sin \theta \right)\]. And then we will apply \[{{\left( a+b+c \right)}^{2}}\] which is equal to \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\].

Complete step-by-step answer:

In this question, we are asked to prove that

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

To prove this expression, we will consider the right-hand side first, so we can write it as
\[RHS=\dfrac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }\]

Now, we will rationalize the denominator by multiplying the numerator and denominator by \[\left[ \left( 1+\cos \theta \right)+\left( \sin \theta \right) \right]\]. So, we will get RHS as
\[RHS=\dfrac{\left( 1+\cos \theta +\sin \theta \right)\left( 1+\cos \theta +\sin \theta \right)}{\left( 1+\cos \theta -\sin \theta \right)\left( 1+\cos \theta +\sin \theta \right)}\]

Now, we know that \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]. So, we can write \[\left( 1+\cos \theta -\sin \theta \right)\left( 1+\cos \theta +\sin \theta \right)\] as \[{{\left( 1+\cos \theta \right)}^{2}}-{{\left( \sin \theta \right)}^{2}}\]. Therefore, we will get RHS as
\[RHS=\dfrac{{{\left( 1+\cos \theta +\sin \theta \right)}^{2}}}{{{\left( 1+\cos \theta \right)}^{2}}-{{\sin }^{2}}\theta }\]

Now, we know that \[{{\left( a+b+c \right)}^{2}}\] can be expressed as \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\] and \[{{\left( a+b \right)}^{2}}\] can be expressed as \[{{a}^{2}}+{{b}^{2}}+2ab\]. Therefore, we will get RHS as

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

Now, we know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] which can also be written as \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]. Therefore, we will write RHS as

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

\[RHS=\dfrac{2+2\cos \theta +2\cos \theta \sin \theta +2\sin \theta }{2{{\cos }^{2}}\theta +2\cos \theta }\]

Now, we know that 2 can be taken out as common from both numerator and denominator. So, we will get,

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

Now, we know that \[\left( \cos \theta \sin \theta +\sin \theta \right)\] can be expressed as \[\left( \cos \theta +1 \right)\sin \theta \] because \[\sin \theta \] is common in both the terms. Also, we know that \[\left( {{\cos }^{2}}\theta +\cos \theta \right)\] can be written as \[\cos \theta \left( \cos \theta +1 \right)\]. Therefore, we will get RHS as,

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

Now, we can see that \[\left( 1+\cos \theta \right)\] can be taken out as common from the numerator. So, we will get RHS as

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

Here, we can see that 2 and \[\left( 1+\cos \theta \right)\] are common in both the numerator and denominator. So, they will cancel out. Therefore, we will get,

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

Now, we will multiply numerator and denominator by \[\cos \theta \]. Therefore, we get
\[RHS=\dfrac{\left( 1+\sin \theta \right)\cos \theta }{{{\cos }^{2}}\theta }\]

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

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

Now, we also know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. So, we will get RHS as,

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

Here, we can see that \[\left( 1+\sin \theta \right)\] is common in both numerator and denominator. Therefore, we can cancel them out. So, we will get RHS as

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

RHS = LHS

Or we can say

LHS = RHS

Hence proved

Note: In this question, there are possibilities that one might get stuck at the point where we show that RHS is equal to \[\dfrac{1+\sin \theta }{\cos \theta }\]. So, after that what we could have done is we would have simplified LHS by rationalizing and proving that \[LHS=\dfrac{1+\sin \theta }{\cos \theta }\]. And in this way, we could have also proved the result.