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Question: If \[\sqrt {\dfrac{{1 - \sin A}}{{1 - \sin A}}} = \dfrac{{\sin A}}{{\cos A}} = \dfrac{1}{{\cos A}}\]...

If 1−sin⁡A1−sin⁡A=sin⁡Acos⁡A=1cos⁡A\sqrt {\dfrac{{1 - \sin A}}{{1 - \sin A}}} = \dfrac{{\sin A}}{{\cos A}} = \dfrac{1}{{\cos A}}, for all permissible values of A{\mathbf{A}}, then A{\mathbf{A}}belongs to
(A) First Quadrant
(B) Second Quadrant
(C) Third Quadrant
(D) Fourth Quadrant

Explanation

Solution

Trigonometric functions are used in this problem. So, firstly, we are to solve the left hand side separately and then we need to compare it with the right hand side of the given problem.
Now, we know that all the Trigonometric ratios are positive in 1st{1^{st}}quadrant, only Sin⁡θ\operatorname{Sin} \theta and Cos⁡ecθ\operatorname{Cos} ec\theta are positive in IIIIquadrant, only Tan⁡θ\operatorname{Tan} \theta and CotθCot\theta are positive in IIIIII quadrant, only Cos⁡θ\operatorname{Cos} \theta & Sec⁡θ\operatorname{Sec} \theta are positive in IVIV quadrant.

Complete step-by-step answer:
As, we are already given that
1−sin⁡A1+sin⁡A+sin⁡Acos⁡A=1cos⁡A\sqrt {\dfrac{{1 - \sin A}}{{1 + \sin A}}} + \dfrac{{\sin A}}{{\cos A}} = \dfrac{1}{{\cos A}}
i.e. 1−sin⁡A1+sin⁡A=1cos⁡A−sin⁡Acos⁡A\sqrt {\dfrac{{1 - \sin A}}{{1 + \sin A}}} = \dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}
Step 2: Now, consider the left hand side only i.e.
1−sin⁡A1+sin⁡A\sqrt {\dfrac{{1 - \sin A}}{{1 + \sin A}}}
Step 3: On multiplying 1−sin⁡A\sqrt {1 - \sin A} with the numerator as well as denominator, we get
(1−sin⁡A)2(1+sin⁡A)(1−Sin⁡A)\sqrt {\dfrac{{{{(1 - \sin A)}^2}}}{{(1 + \sin A)(1 - \operatorname{Sin} A)}}}
Step 4: On using the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}we get..
⇒\Rightarrow (1−sin⁡A)21−sin⁡A=(1−sin⁡A)2cos⁡2A\sqrt {\dfrac{{{{(1 - \sin A)}^2}}}{{1 - \sin A}}} = \sqrt {\dfrac{{{{(1 - \sin A)}^2}}}{{{{\cos }^2}A}}}
because 1−sin⁡2A=cos⁡2A1 - {\sin ^2}A = {\cos ^2}A
Step 5: Rearrange & solve the terms..
⇒\Rightarrow (1cos⁡A−sin⁡Acos⁡A)2=∣1cos⁡A=sin⁡Acos⁡A∣\sqrt {{{\left( {\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}} \right)}^2}} = \left| {\dfrac{1}{{\cos A}} = \dfrac{{\sin A}}{{\cos A}}} \right|
Whereas the given right hand side is
⇒\Rightarrow 1cos⁡A=sin⁡Acos⁡A\dfrac{1}{{\cos A}} = \dfrac{{\sin A}}{{\cos A}}
Hence
⇒\Rightarrow ∣1cos⁡A=sin⁡Acos⁡A∣=1cos⁡A=sin⁡Acos⁡A\left| {\dfrac{1}{{\cos A}} = \dfrac{{\sin A}}{{\cos A}}} \right| = \dfrac{1}{{\cos A}} = \dfrac{{\sin A}}{{\cos A}}
Consider, (1cos⁡A−sin⁡Acos⁡A)=y\left( {\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}} \right) = y
Therefore, ∣y∣=y\left| y \right| = y which is possible if and only if y⩾0y \geqslant 0
Hence, 1cos⁡A−sin⁡Acos⁡A⩾0\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}} \geqslant 0
1cos⁡A(1−sin⁡A)⩾0\dfrac{1}{{\cos A}}\left( {1 - \sin A} \right) \geqslant 0
Which is possible if cos⁡A\cos A is positive. Which is possible if and only if AA lies either in II quad or IVIV quad.

Note: There are 44 quadrants in the Trigonometry. In which all the trigonometric ratios are positive in the 1st{1^{st}} one whereas pairs of trigonometric ratios are positive in the rest of the quadrants.
In the IIquad, θ\theta lies between 00 to π2\dfrac{\pi }{2}. In the IIIIquad, θ\theta lies between π2\dfrac{\pi }{2} to π\pi . In the IIIIII quad, θ\theta lies between π\pi to 3π2\dfrac{{3\pi }}{2}. In the IVIVquad, θ\theta lies between 3π2\dfrac{{3\pi }}{2}to2π2\pi .
Modulus of xx will be equal to xxitself, if and only if xx is positive or zero. Hence the correct options are (A)(A) and(B)(B).