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

Prove that: cos⁡A1−sin⁡A=1+sin⁡Acos⁡A\dfrac{{\cos A}}{{1 - \sin A}} = \dfrac{{1 + \sin A}}{{\cos A}}.

Explanation

Solution

To solve the question, we will start with the LHS and try to get the RHS from it. We can see that in the denominator of RHS there is a cosine term and in the denominator of LHS it is the sine term. So, to get the cosine term, we will multiply the numerator and denominator of LHS by 1+sin⁡A1 + \sin A. Then, with the use of trigonometry we will solve further to get the RHS.

Complete step-by-step solution:
LHS: cos⁡A1−sin⁡A\dfrac{{\cos A}}{{1 - \sin A}}
Now in order to get the cosine term in the denominator we will multiply both the numerator and denominator by 1+sin⁡A1 + \sin A. So, we get;
=cos⁡A1−sin⁡A×1+sin⁡A1+sin⁡A= \dfrac{{\cos A}}{{1 - \sin A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}
Now we can see that the denominator is in the form (a+b)(a−b)\left( {a + b} \right)\left( {a - b} \right) and we know that:
(a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
Therefore,
(1−sin⁡A)(1+sin⁡A)=1−sin⁡2A\left( {1 - \sin A} \right)\left( {1 + \sin A} \right) = 1 - {\sin ^2}A
So, we get;
=cos⁡A(1+sin⁡A)1−sin⁡2A= \dfrac{{\cos A\left( {1 + \sin A} \right)}}{{1 - {{\sin }^2}A}}
Now we know that, sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
∴1−sin⁡2A=cos⁡2A\therefore 1 - {\sin ^2}A = {\cos ^2}A
Now we will use this relation in the above fraction. So, we get;
=cos⁡A(1+sin⁡A)cos⁡2A= \dfrac{{\cos A\left( {1 + \sin A} \right)}}{{{{\cos }^2}A}}
Now we will cancel the cosine term from both the numerator and denominator. So, we get,
=(1+sin⁡A)cos⁡A= \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}
And this is equal to the RHS.

Note: In this question one can also think of starting with the RHS and try to get the LHS in the process. But either that will be a very lengthy approach or one can not arrive at the result.
We can also solve this by another simple method that is by dividing the LHS by cos⁡A\cos A and applying the formula that: sec⁡2A−tan⁡2A=1{\sec ^2}A - {\tan ^2}A = 1.
LHS: cos⁡A1−sin⁡A\dfrac{{\cos A}}{{1 - \sin A}}
Dividing the numerator and denominator by cos⁡A\cos A.
=(cos⁡Acos⁡A)1−sin⁡Acos⁡A= \dfrac{{\left( {\dfrac{{\cos A}}{{\cos A}}} \right)}}{{\dfrac{{1 - \sin A}}{{\cos A}}}}
Simplifying we get;
=1(1cos⁡A−sin⁡Acos⁡A)= \dfrac{1}{{\left( {\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}} \right)}}
Now we will write 1cos⁡A=sec⁡A\dfrac{1}{{\cos A}} = \sec A and sin⁡Acos⁡A=tan⁡A\dfrac{{\sin A}}{{\cos A}} = \tan A. So, we get;
=1sec⁡A−tan⁡A= \dfrac{1}{{\sec A - \tan A}}
Now we will multiply and divide by sec⁡A+tan⁡A\sec A + \tan A.
=sec⁡A+tan⁡A(sec⁡A+tan⁡A)(sec⁡A−tan⁡A)= \dfrac{{\sec A + \tan A}}{{\left( {\sec A + \tan A} \right)\left( {\sec A - \tan A} \right)}}
On simplification we get;
=sec⁡A+tan⁡Asec⁡2A−tan⁡2A= \dfrac{{\sec A + \tan A}}{{{{\sec }^2}A - {{\tan }^2}A}}
Now we will use sec⁡2A−tan⁡2A=1{\sec ^2}A - {\tan ^2}A = 1. So, we get,
=sec⁡A+tan⁡A= \sec A + \tan A
Writing in the form of sine and cosine. We get,
=1cos⁡A+sin⁡Acos⁡A= \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}
Taking the LCM and solving we get;
=1+sin⁡Acos⁡A= \dfrac{{1 + \sin A}}{{\cos A}}
And this is the RHS.