Given $\dfrac{\pi }{2} < \alpha < \pi $ , then the expression $\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=$
A. $\dfrac{1}{\cos \alpha }$ .
B. $-\dfrac{2}{\cos \alpha }$ .
C. $\dfrac{2}{\cos \alpha }$ .
D. None of these

Hint: In this problem we need to find the value of the given expression. We will first consider each term individually in the given expression and rationalize the denominators by multiplying appropriate factors. Now we will apply different algebraic formulas like $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , $a\times a\times a\times ....\text{ n times}={{a}^{n}}$ . Now we will apply the simplified values in the given expression and simplify them by using exponential formula ${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}$ . After that we will use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to simplify the expression. Now we will perform some basic mathematical operations to simplify the expression and change the sign of trigonometric ratios based on the given quadrant.

Complete step by step answer:
Given that, $\dfrac{\pi }{2} < \alpha < \pi $ and $\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=$
From the value $\dfrac{\pi }{2} < \alpha < \pi $, we can say that the angle $\alpha $ lies in the second quadrant. We know that the trigonometric ratios $\sin $ , $\csc $ are positive while the remaining are negative.
Consider the term $\dfrac{1-\sin \alpha }{1+\sin \alpha }$ from the given expression.
Rationalizing the above term by multiplying and dividing with $1-\sin \alpha $ , then we will get
$\dfrac{1-\sin \alpha }{1+\sin \alpha }=\dfrac{1-\sin \alpha }{1+\sin \alpha }\times \dfrac{1-\sin \alpha }{1-\sin \alpha }$
Using the exponential formulas $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , $a\times a\times a\times ....\text{ n times}={{a}^{n}}$ in the above equation, then we will have
$\begin{align}
  & \dfrac{1-\sin \alpha }{1+\sin \alpha }=\dfrac{{{\left( 1-\sin \alpha \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\alpha } \\
 & \Rightarrow \dfrac{1-\sin \alpha }{1+\sin \alpha }=\dfrac{{{\left( 1-\sin \alpha \right)}^{2}}}{1-{{\sin }^{2}}\alpha } \\
\end{align}$
Applying the simplified terms in the given expression, then we will get
$\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\sqrt{\dfrac{{{\left( 1-\sin \alpha \right)}^{2}}}{1-{{\sin }^{2}}\alpha }}+\sqrt{\dfrac{{{\left( 1+\sin \alpha \right)}^{2}}}{1-{{\sin }^{2}}\alpha }}$
From the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$we can write $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $ . Substituting this value in the above equation, then we will have
$\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\sqrt{\dfrac{{{\left( 1-\sin \alpha \right)}^{2}}}{{{\cos }^{2}}\alpha }}+\sqrt{\dfrac{{{\left( 1+\sin \alpha \right)}^{2}}}{{{\cos }^{2}}\alpha }}$
Applying the exponential formula ${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}$ in the above equation, then we will get
$\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\dfrac{\sqrt{{{\left( 1-\sin \alpha \right)}^{2}}}}{\sqrt{{{\cos }^{2}}\alpha }}+\dfrac{\sqrt{{{\left( 1+\sin \alpha \right)}^{2}}}}{\sqrt{{{\cos }^{2}}\alpha }}$
We know that both square and square root will get cancelled to each other, then we will have
$\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\dfrac{\left| 1-\sin \alpha \right|}{\left| \cos \alpha \right|}+\dfrac{\left| 1+\sin \alpha \right|}{\left| \cos \alpha \right|}$
We know that the angle $\alpha $ lies in the second quadrant. So the value of $\sin $ is positive and the value of $\cos $ is negative. Applying these signs in the above equation, then we will get
$\sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\dfrac{1-\sin \alpha }{-\cos \alpha }+\dfrac{1+\sin \alpha }{-\cos \alpha }$
Performing basic mathematical operations in the above equation, then we will have
$\begin{align}
  & \sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=\dfrac{1-\sin \alpha +1+\sin \alpha }{-\cos \alpha } \\
 & \Rightarrow \sqrt{\dfrac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\dfrac{1+\sin \alpha }{1-\sin \alpha }}=-\dfrac{2}{\cos \alpha } \\
\end{align}$

So, the correct answer is “Option B”.

Note: While solving this problem we need to consider the sign of the trigonometric ratios. So we have considered a modulus function after canceling square and square root. It is must and should when the quadrant of the angle is mentioned otherwise it is not necessary.