How do you simplify the given sum: $\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$?

Hint: We start solving the problem by equating the given sum to a variable. We then make use of the fact that $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$, $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ and $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ to proceed through the problem. We then make the necessary calculations and then make use of the fact that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $ to proceed further through the problem. We then make the necessary calculations and make use of the fact that $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $\sec \theta =\dfrac{1}{\cos \theta }$ and $\dfrac{1}{\sin x}=\operatorname{cosec}x$ to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the result of the given sum: $\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$.
Let us assume $d=\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$ ---(1).
We know that $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$, $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ and $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$. Let us use these results in equation (2).
$\Rightarrow d=\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}+\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\sin }^{2}}\dfrac{x}{2}}$.
$\Rightarrow d=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}$ ---(2).
We know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $. Let us use these results in equation (2).
$\Rightarrow d=\tan \dfrac{x}{2}+\cot \dfrac{x}{2}$ ---(3).
We know that $\cot \theta =\dfrac{1}{\tan \theta }$. Let us use this result in equation (3).
$\Rightarrow d=\tan \dfrac{x}{2}+\dfrac{1}{\tan \dfrac{x}{2}}$.
$\Rightarrow d=\dfrac{{{\tan }^{2}}\dfrac{x}{2}+1}{\tan \dfrac{x}{2}}$ ---(4).
We know that $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$. Let us use this result in equation (4).
$\Rightarrow d=\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{\tan \dfrac{x}{2}}$ ---(5).
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$. Let us use these results in equation (5).
$\Rightarrow d=\dfrac{\dfrac{1}{{{\cos }^{2}}\dfrac{x}{2}}}{\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}$.
$\Rightarrow d=\dfrac{1}{\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$.
$\Rightarrow d=\dfrac{2}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$ ---(6).
We know that $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$. Let us use this result in equation (6).
$\Rightarrow d=\dfrac{2}{\sin x}$ ---(7).
We know that $\dfrac{1}{\sin x}=\operatorname{cosec}x$. Let us use this result in equation (7).
$\Rightarrow d=2\operatorname{cosec}x$.
So, we have found the simplified form of the given sum $\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$ as $2\operatorname{cosec}x$.
$\therefore $ The simplified form of the given sum $\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$ is $2\operatorname{cosec}x$.

Note:
We can also solve the given problem as shown below:
We have given $d=\dfrac{\sin x}{1+\cos x}+\dfrac{\sin x}{1-\cos x}$.
$\Rightarrow d=\dfrac{\sin x\left( 1-\cos x \right)+\sin x\left( 1+\cos x \right)}{\left( 1+\cos x \right)\left( 1-\cos x \right)}$.
$\Rightarrow d=\dfrac{\sin x-\sin x\cos x+\sin x+\sin x\cos x}{1-\cos x+\cos x-{{\cos }^{2}}x}$.
$\Rightarrow d=\dfrac{2\sin x}{1-{{\cos }^{2}}x}$ ---(8).
We know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$. Let us use this result in equation (8).
$\Rightarrow d=\dfrac{2\sin x}{{{\sin }^{2}}x}$.
$\Rightarrow d=\dfrac{2}{\sin x}$ ---(9).
We know that $\dfrac{1}{\sin x}=\operatorname{cosec}x$. Let us use this result in equation (9).
$\Rightarrow d=2\operatorname{cosec}x$.