Question

If $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \cos ecA = \cot A$, then ${\mathbf{A}}$, ${\mathbf{A}}$lies in the quadrants.
(A) I, II
(B) II, III
(C) I, III
(D) I, IV

Answer 1

Trigonometric functions are used in this problem. So, firstly we all solve the left-hand side separately, and then we need to compare it to the right-hand side of the given problem.
Now, we know that all the Trigonometric ratios are positive in ${1^{st}}$a quadrant, only $\operatorname{sin} \theta$ and $\operatorname{cos} ec\theta$ are positive in $II$the quadrant, only $\operatorname{tan} \theta$ and $cot\theta$ are positive in $III$ the quadrant, only $\operatorname{Cos} \theta$ & $\operatorname{sec} \theta$ are positive in $IV$ the quadrant.

Complete step-by-step answer:
Step 1: Take the left had the side of the given
Problem i.e.
$\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}}$
Step 2: Multiply the numerator & denominator $(1 + \cos A)$inside the square root.
i.e. $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{(1 - \cos A)(1 + \cos A)}}}$
Step 3: On solving the denominator using the identity $(a - b)(a + b) = {a^2} - {b^2}$ , we get.
$\Rightarrow$ $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{1 - {{\cos }^2}A}}}$
Step 4: We know that, $1 - {\cos ^2}A - {\sin ^2}A$
Because, ${\sin ^2}A - {\cos ^2}A = 1$
On putting the value of ${\sin ^2}A$in the, we get
$\Rightarrow$ $\sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{{{\sin }^2}A}}}$
Step 5: On rearranging the terms, we get
$\Rightarrow$ $\sqrt {\dfrac{{{{\left( {1 + \cos A} \right)}^2}}}{{1 - \cos A}}} = \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{{{\sin }^2}A}}}$
Step 6: We know that, $\dfrac{1}{{\sin \theta }} = \cos ec\theta$
and $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$
On putting these values, we get
$\Rightarrow$ $\dfrac{{\sqrt {1 + \cos A} }}{{1 - \cos A}} = \sqrt {{{\left( {\cos ecA + \cot A} \right)}^2}}$
$= \left| {\cos ecA + \cot A} \right|$
Step 7: Now according to us,
$\Rightarrow$ $\dfrac{{\sqrt {1 + \cos A} }}{{1 - \cos A}} = \left| {\cos ecA + \cot A} \right|$
But, according to the given problem,
$\Rightarrow$ $\dfrac{{\sqrt {1 + \cos A} }}{{1 - \cos A}} = \cos ecA + \cot A$
Step 8: Therefore,
$\Rightarrow$ $\left| {\cos ecA + \cot A} \right| = \cos ecA + \cot A$
Considering $\cos ecA + \cot A = x$
i.e. $\left| x \right| = x$ if and only if $x \geqslant 0$
Step 9: That means,
$\Rightarrow$ $x = \cos ecA + \cot A \geqslant 0$
This implies, $\sin A > 0$
Which is possible if $A$ lies either in $I$quadrant or $II$quadrant.

Note: There are $4$ quadrants in the Trigonometry. In which all the trigonometric ratios are positive in the ${I^{st}}$ 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 $(A)$ $I,II$quadrant.