Question

If \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A\], then the quadrant in which the angle $A$ lies are?
A)$I,II$
B)$II,III$
C)$I,IV$
D)$III,IV$

Answer 1

Hint:
We are asked to find the quadrant in which $A$ belongs for the equation to be true. For that we have to start with the left hand side and check for all values of $A$, this expression is equal to the right hand side. Once we find the favourable values of $A$, we can find the quadrants.

Useful formula:
For any $a,b$ we have,
$(a + b)(a - b) = {a^2} - {b^2}$
For any angle $\theta $ we have,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
A rectangular coordinate system is divided into four quadrants.
Angles between $0$ and ${90^ \circ }$ belong to the first quadrant.
Angles between ${90^ \circ }$ and ${180^ \circ }$ belong to the second quadrant.
Angles between $180^\circ $ and ${270^ \circ }$ belong to the third quadrant.
Angles between $270^\circ $ and ${360^ \circ }$ belong to the fourth quadrant.
For every $x$, we have $\sqrt {{x^2}} = x$ if and only if $x \geqslant 0$.

Complete step by step solution:
Given that, \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A\]
We need to find which quadrant does $A$ belong.
Consider \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A\]
We can multiply and divide the expression inside the root in the left hand side by $1 + \sin A$.
This gives,
\[\sqrt {\dfrac{{(1 + \sin A)(1 + \sin A)}}{{(1 - \sin A)(1 + \sin A)}}} = \sec A + \tan A\]
 Now the denominator inside the root is in the form $(a + b)(a - b)$.
For any $a,b$ we have,
$(a + b)(a - b) = {a^2} - {b^2}$
Also, the numerator inside the root can be written as ${(1 + \sin A)^2}$.
Substituting these we have,
\[\sqrt {\dfrac{{{{(1 + \sin A)}^2}}}{{1 - {{\sin }^2}A}}} = \sec A + \tan A\]

For any angle $\theta $ we have,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
This gives $ \Rightarrow 1 - {\sin ^2}\theta = {\cos ^2}\theta $

Using this we can substitute in the denominator of the above equation.
\[\sqrt {\dfrac{{{{(1 + \sin A)}^2}}}{{{{\cos }^2}A}}} = \sec A + \tan A\]
\[ \Rightarrow \sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \sec A + \tan A\]

Now we have, $\sqrt {{x^2}} = x$ if and only if $x \geqslant 0$.
So, \[\sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \dfrac{{1 + \sin A}}{{\cos A}}\] if and only if \[\dfrac{{1 + \sin A}}{{\cos A}} \geqslant 0\]

If \[\sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \dfrac{{1 + \sin A}}{{\cos A}}\], then
\[\sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}\]
\[ \Rightarrow \sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \sec A + \tan A\], which is our result.
So this result holds when \[\sqrt {{{(\dfrac{{1 + \sin {A^{}}}}{{{{\cos }^{}}A}})}^2}} = \dfrac{{1 + \sin A}}{{\cos A}}\], that is when,\[\dfrac{{1 + \sin A}}{{\cos A}} \geqslant 0\]
We have $\sin A,\cos A \in [ - 1,1]$
$\therefore 1 + \sin A \geqslant 0$
But $\cos A$ can be negative as well.
So, we have to eliminate the values of $A$ for which $\cos A$ is negative.
$\cos A < 0,A \in (90^\circ ,270^\circ )$, which are the second and third quadrant.
So, for every $A \in (0,{90^ \circ }) \cup (270^\circ ,{360^ \circ })$, \[\dfrac{{1 + \sin A}}{{\cos A}} \geqslant 0\]
This means \[\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A\] holds if, $A \in (0,{90^ \circ }) \cup (270^\circ ,{360^ \circ })$ the first and fourth quadrant.

$\therefore $ The answer is option C.

Note:
Here we find the favourable values of $A$ by considering where $1 + \sin A$ and $\cos A$ are positive. If the expression inside the root symbol was positive always then we can simplify $\sqrt {{x^2}} $ to $x$ directly. If so, the equation holds in every quadrant.