Question

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

Answer 1

Hint: 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 ${1^{st}}$quadrant, only $\operatorname{Sin} \theta $ and $\operatorname{Cos} ec\theta $ are positive in $II$quadrant, only $\operatorname{Tan} \theta $ and $Cot\theta $ are positive in $III$ quadrant, only $\operatorname{Cos} \theta $ & $\operatorname{Sec} \theta $ are positive in $IV$ quadrant.

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

Note: There are $4$ quadrants in the Trigonometry. In which all the trigonometric ratios are positive in the ${1^{st}}$ one whereas pairs of trigonometric ratios are positive in the rest of the quadrants.
In the $I$quad, $\theta $ lies between $0$ to $\dfrac{\pi }{2}$. In the $II$quad, $\theta $lies between $\dfrac{\pi }{2}$ to $\pi $. In the $III$ quad, $\theta $ lies between $\pi $ to $\dfrac{{3\pi }}{2}$. In the $IV$quad, $\theta $ lies between $\dfrac{{3\pi }}{2}$to$2\pi $.
Modulus of $x$ will be equal to $x$itself, if and only if $x$ is positive or zero. Hence the correct options are $(A)$ and$(B)$.