Question

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

Answer 1

Hint: In this question we are given \[\dfrac{{\cos \theta }}{{1 + \sin \theta }}\]and we have to solve to make it equal to \[\dfrac{{1 - \sin \theta }}{{\cos \theta }}\]. So here firstly we will start with LHS and will multiply and divide the numerator by \[1 - \sin \theta \]. After getting the desired equation now we will use identity \[{\sin ^2}x = 1 - {\cos ^2}x\]and \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]for solving further and by cancelling the common terms we get desired results.

Formula used:
By using the rationalization and elimination method, and rules that \[[(a + b)(a - b) = {a^2} - {b^2}]\]and \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]. we get the desired result

Complete step-by-step answer:
Here we are asked to prove that \[\dfrac{{\cos \theta }}{{1 + \sin \theta }} = \dfrac{{1 - \sin \theta }}{{\cos \theta }}\]
So we will start with LHS and on LHS we have \[\dfrac{{\cos \theta }}{{1 + \sin \theta }}\]
We will multiply and divide the numerator and denominator by \[1 - \sin \theta \] so we get the equation as
\[\dfrac{{\cos \theta }}{{1 + \sin \theta }} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}\]
Now as we know that \[[(a + b)(a - b) = {a^2} - {b^2}]\]
So on further simplification we get the follow equation
\[\dfrac{{\cos \theta (1 - \sin \theta )}}{{1 - {{\sin }^2}\theta }}\]
Now by using the identity \[{\sin ^2}x = 1 - {\cos ^2}x\]and \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] the equation we obtained is
\[\dfrac{{\cos \theta (1 - \sin \theta )}}{{{{\cos }^2}\theta }}\]
Later on by cancelling the common terms and after further simplification we get the desired equation which is
\[\dfrac{{1 - \sin \theta }}{{\cos \theta }}\]
Here we will observe that now on LHS we have \[\dfrac{{1 - \sin \theta }}{{\cos \theta }}\]and on RHS we have\[\dfrac{{1 - \sin \theta }}{{\cos \theta }}\]which means \[LHS = RHS\].
Hence the above given question is proved.

Note: While solving such types of questions we need to remember that\[{\sin ^2}x = 1 - {\cos ^2}x\]. The most important identity while solving such types of questions is \[[(a + b)(a - b) = {a^2} - {b^2}]\]and the trigonometric identity used for solving above question is \[{\cos ^2}\theta + {\sin ^2}\theta = 1\].Keep in mind that while simplifying such questions we need to very careful because if we get missed in the initial stage we will get wrong solution.