Solve ${{\tan }^{-1}}\left[ \dfrac{\cos x}{1+\sin x} \right]$.

Hint: For solving this question you should know about the general formulas of trigonometry for 2a angles. In this problem we will simply divide the angle $x$ as $\dfrac{2x}{2}$ and thus it will be in the form of angle 2a. Then we will use the formulas of $\cos 2a$ and $\sin 2a$ to solve this problem. Then we divide it by $\cos \dfrac{x}{2}$ and solve it forward.

Complete step by step answer:
According to the problem, we have to solve the expression ${{\tan }^{-1}}\left[ \dfrac{\cos x}{1+\sin x} \right]$. We know that,
$\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$
Replacing $x$ by $\dfrac{x}{2}$, we get,
$\begin{align}
  & \cos \left( \dfrac{2x}{2} \right)={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\
 & \Rightarrow \cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
We also know that,
$\sin 2x=2\sin x\cos x$
Replacing $x$ by $\dfrac{x}{2}$, we get,
$\begin{align}
  & \sin \left( \dfrac{2x}{2} \right)=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
 & \Rightarrow \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
From equations (i) and (ii), we get,
$\begin{align}
  & {{\tan }^{-1}}\left[ \dfrac{\cos x}{1+\sin x} \right]={{\tan }^{-1}}\left[ \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{1+\left( 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right)} \right] \\
 & \Rightarrow {{\tan }^{-1}}\left[ \dfrac{\cos x}{1+\sin x} \right]={{\tan }^{-1}}\left[ \dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{1+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right] \\
\end{align}$
As ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Replacing $x$ by $\dfrac{x}{2}$, we get,
${{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}=1$
So, we can also write it as:
\[\begin{align}
  & ={{\tan }^{-1}}\left[ \dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}+\sin \dfrac{x}{2}} \right] \\
\end{align}\]
Dividing the numerator and denominator by $\cos \dfrac{x}{2}$ we get,
\[\begin{align}
  & ={{\tan }^{-1}}\left[ \dfrac{\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{1-\tan \dfrac{x}{2}}{1+\tan \dfrac{x}{2}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}} \right] \\
\end{align}\]
Using $\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$, we get,
$\begin{align}
  & ={{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right] \\
 & =\dfrac{\pi }{4}-\dfrac{x}{2} \\
\end{align}$
So, the final answer is $\dfrac{\pi }{4}-\dfrac{x}{2}$.

Note: While solving these type of questions you have to be careful about the formulas of $\tan \left( x\pm y \right)$ or $\sin 2x$ and $\cos 2x$, because if any one of them will be wrong or any sign will be wrong, then we can solve the questions completely, but the answer will be totally wrong.