If a function is given as $y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$ then $\dfrac{dy}{dx}$ equals:
A.$\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$
B.$\dfrac{1}{2}{{\csc }^{2}}\dfrac{x}{2}$
C.${{\sec }^{2}}\dfrac{x}{2}$
D.${{\csc }^{2}}\dfrac{x}{2}$

Hint: It is given that $y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$. We know that $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ so substituting the values of $\left( 1-\cos x \right)\And \left( 1+\cos x \right)$ in equation of y we get, $y=\tan \dfrac{x}{2}$. Now, differentiate y with respect to x which will give the required answer. The differentiation is done using chain rule.

Complete step-by-step answer:
It is given that:
$y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$
We know that,
$1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$
$1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$
Substituting the above values of $\left( 1-\cos x \right)\And \left( 1+\cos x \right)$ in equation of y we get,
$\begin{align}
  & y=\sqrt{\dfrac{1-\cos x}{1+\cos x}} \\
 & \Rightarrow y=\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}} \\
\end{align}$
In the above equation, 2 will be cancelled out from the numerator and the denominator.
$\begin{align}
  & y=\sqrt{\dfrac{{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}}} \\
 & \Rightarrow y=\sqrt{{{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}}.........Eq.\left( 1 \right) \\
\end{align}$
We know that,
$\tan \dfrac{x}{2}=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}$
Substituting the above relation in eq. (1) we get,
$\begin{align}
  & y=\sqrt{{{\left( \tan \dfrac{x}{2} \right)}^{2}}} \\
 & \Rightarrow y=\tan \dfrac{x}{2}.........Eq.\left( 2 \right) \\
\end{align}$
Let us assume that;
 $\dfrac{x}{2}=t$
On cross – multiplying the above equation we get,
$x=2t$
Differentiating on both the sides we get,
$dx=2dt$……….. Eq. (3)
Substituting $\dfrac{x}{2}=t$ in eq. (2) we get,
$y=\tan t$
Differentiating both the sides with respect to x we get,
$\dfrac{dy}{dt}={{\sec }^{2}}t$……. Eq. (4)
As we have to find $\dfrac{dy}{dx}$ so we can write $\dfrac{dy}{dt}$ as follows:
$\dfrac{dy}{dt}=\dfrac{dy}{dx}\left( \dfrac{dx}{dt} \right)$…… Eq. (5)
Rearranging eq. (3) we get,
$\begin{align}
  & dx=2dt \\
 & \Rightarrow \dfrac{dx}{dt}=2 \\
\end{align}$
Substituting the above value in eq. (5) we get,
$\dfrac{dy}{dt}=\dfrac{dy}{dx}\left( 2 \right)$
Now, substituting the above value in eq. (4) and $t=\dfrac{x}{2}$ we get,
$2\dfrac{dy}{dx}={{\sec }^{2}}\dfrac{x}{2}$
Dividing 2 on both the sides of the above equation we get,
$\dfrac{dy}{dx}=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$
From the above solution, we have got the derivative of $y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$ with respect to x is $\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$.
Hence, the correct option is (a).

Note: The plausible areas in the solution where you can go wrong is:
In writing the trigonometric identity of $\left( 1-\cos x \right)\And \left( 1+\cos x \right)$, you might interchange the values corresponding to these trigonometric functions which is demonstrated below.
$1-\cos x=2{{\cos }^{2}}\dfrac{x}{2}$
$1+\cos x=2{{\sin }^{2}}\dfrac{x}{2}$
Be careful about where 2 should come in the result of the derivative with respect to x. There is a chance of calculation mistake in the final answer whether 2 should be multiplied or divided by ${{\sec }^{2}}\dfrac{x}{2}$.