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Question: The value of \[\dfrac{{1 - \sin A}}{{\cos A}}\] is equal to (A) \[\dfrac{{\cos A}}{{1 + \sin A}}\]...

The value of 1−sin⁡Acos⁡A\dfrac{{1 - \sin A}}{{\cos A}} is equal to
(A) cos⁡A1+sin⁡A\dfrac{{\cos A}}{{1 + \sin A}}
(B) sin⁡A1−cos⁡A\dfrac{{\sin A}}{{1 - \cos A}}
(C) tan⁡A1+tan⁡A\dfrac{{\tan A}}{{1 + \tan A}}
(D) tan⁡A1+cos⁡A\dfrac{{\tan A}}{{1 + \cos A}}

Explanation

Solution

In the given question, we have two simple trigonometric functions, sine and cosine. We do not have direct formulae to apply and simplify. So, we make modifications in the numerator and the denominator by multiplying and dividing a particular term. By doing this, we can simplify it into a form where we can cancel out or group terms so that we can apply formulae easily. Then, we can simplify the terms and use the available formulae to get to the final answer.

Formula used:
(a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1

Complete step by step answer:
Let the consider the given expression,
1−sin⁡Acos⁡A\dfrac{{1 - \sin A}}{{\cos A}}
We have 1−sin⁡A1 - \sin A in the numerator. In order to use the (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} formula, we need 1+sin⁡A1 + \sin A. So, we multiply and divide by 1+sin⁡A1 + \sin A.
⇒1−sin⁡Acos⁡A=1−sin⁡Acos⁡A×1+sin⁡A1+sin⁡A\Rightarrow \dfrac{{1 - \sin A}}{{\cos A}} = \dfrac{{1 - \sin A}}{{\cos A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}
Let us now group the numerators and the denominators,
⇒1−sin⁡Acos⁡A=(1−sin⁡A)(1+sin⁡A)cos⁡A(1+sin⁡A)\Rightarrow \dfrac{{1 - \sin A}}{{\cos A}} = \dfrac{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}}{{\cos A\left( {1 + \sin A} \right)}}…… (1)
We can see that the numerator is in the form (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
Where,
a=1a = 1
b=sin⁡Ab = \sin A
⇒(1−sin⁡A)(1+sin⁡A)=1−sin⁡2A\Rightarrow \left( {1 - \sin A} \right)\left( {1 + \sin A} \right) = 1 - {\sin ^2}A
Substituting in (1), we get
⇒1−sin⁡Acos⁡A=1−sin⁡2Acos⁡A(1+sin⁡A)\Rightarrow \dfrac{{1 - \sin A}}{{\cos A}} = \dfrac{{1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}
We know that,
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
⇒1−sin⁡2A=cos⁡2A\Rightarrow 1 - {\sin ^2}A = {\cos ^2}A
Substituting this, we get
⇒1−sin⁡Acos⁡A=cos⁡2Acos⁡A(1+sin⁡A)\Rightarrow \dfrac{{1 - \sin A}}{{\cos A}} = \dfrac{{{{\cos }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}
We can cancel out the cos⁡A\cos A in the denominator.
Now, we get
⇒1−sin⁡Acos⁡A=cos⁡A1+sin⁡A\Rightarrow \dfrac{{1 - \sin A}}{{\cos A}} = \dfrac{{\cos A}}{{1 + \sin A}}
Therefore, the final answer is cos⁡A1+sin⁡A\dfrac{{\cos A}}{{1 + \sin A}}. Hence, option (A) is the correct answer.

Note:
The question does not give us an expression where we can apply a formula directly. So, we need to make changes according to the terms present. The denominator cannot be simplified since it's only a trigonometric function. So, we can only change the numerator. Note that the question contains trigonometric functions, but we also use algebraic formulae. So, we need to be thorough with all the formulae.