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

Prove thatcos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }} = \dfrac{{1 - \sin \theta }}{{\cos \theta }}.

Explanation

Solution

In this question we are given cos⁡θ1+sin⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }}and we have to solve to make it equal to 1−sin⁡θcos⁡θ\dfrac{{1 - \sin \theta }}{{\cos \theta }}. So here firstly we will start with LHS and will multiply and divide the numerator by 1−sin⁡θ1 - \sin \theta . After getting the desired equation now we will use identity sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}xand cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1for 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)=a2−b2][(a + b)(a - b) = {a^2} - {b^2}]and cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1. we get the desired result

Complete step-by-step answer:
Here we are asked to prove that cos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }} = \dfrac{{1 - \sin \theta }}{{\cos \theta }}
So we will start with LHS and on LHS we have cos⁡θ1+sin⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }}
We will multiply and divide the numerator and denominator by 1−sin⁡θ1 - \sin \theta so we get the equation as
cos⁡θ1+sin⁡θ×1−sin⁡θ1−sin⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}
Now as we know that [(a+b)(a−b)=a2−b2][(a + b)(a - b) = {a^2} - {b^2}]
So on further simplification we get the follow equation
cos⁡θ(1−sin⁡θ)1−sin⁡2θ\dfrac{{\cos \theta (1 - \sin \theta )}}{{1 - {{\sin }^2}\theta }}
Now by using the identity sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}xand cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 the equation we obtained is
cos⁡θ(1−sin⁡θ)cos⁡2θ\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
1−sin⁡θcos⁡θ\dfrac{{1 - \sin \theta }}{{\cos \theta }}
Here we will observe that now on LHS we have 1−sin⁡θcos⁡θ\dfrac{{1 - \sin \theta }}{{\cos \theta }}and on RHS we have1−sin⁡θcos⁡θ\dfrac{{1 - \sin \theta }}{{\cos \theta }}which means LHS=RHSLHS = RHS.
Hence the above given question is proved.

Note: While solving such types of questions we need to remember thatsin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x. The most important identity while solving such types of questions is [(a+b)(a−b)=a2−b2][(a + b)(a - b) = {a^2} - {b^2}]and the trigonometric identity used for solving above question is cos⁡2θ+sin⁡2θ=1{\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.