Prove the given inverse trigonometric expression, \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2},x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\] .

Hint: In this question, we have the inverse tan function, but we don’t have a tan function inside that inverse function. So, first, we need to convert that part in terms of tan. Using the formula \[\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\] , \[\sin x=2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}\] and the identity \[1={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}\]. Put these all, and solve further.

Complete step-by-step solution -
According to the question, we have
\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)\]……………(1)
In equation (1), we have the tan inverse of \[\dfrac{\cos x}{1+\sin x}\].
Here we have to remove this inverse of tan. For removing we have to make the term \[\dfrac{\cos x}{1+\sin x}\] in the form of tan.
Let’s proceed with the numerator.
We know that,
\[\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\]………….(2)
We also know that,
\[\sin x=2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}\]……………..(3)
\[1={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}\]……………..(4)
Adding equation (3) and equation (4), we get
\[1+\sin x={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}\] ………….(5)
From equation (1), we have \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)\] .
Putting equation (2) and equation (5) in equation (1), we get
\[{{\tan }^{-1}}\left( \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2\cos \dfrac{x}{2}.\sin \dfrac{x}{2}} \right)\] ……………(6)
We also know that,
\[{{A}^{2}}-{{B}^{2}}=\left( A+B \right).\text{ }\left( A-B \right)\] .
Similarily equation (2) can be written as,
\[\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right).\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)\] …………..(7)
And also,
\[{{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2.A.B\] .
Similarily equation (5) can be written as,
\[{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}={{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}\]…………….(8)
Using equation (7) and equation (8), we can write equation (6) as,
\[{{\tan }^{-1}}\left( \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}} \right)\]
\[\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\] is common in the numerator as well as the denominator. So, we can cancel one \[\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\] in both numerator and denominator. Our equation will look like, \[{{\tan }^{-1}}\left( \dfrac{\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}+\sin \dfrac{x}{2}} \right)\] .
Dividing by \[\cos \dfrac{x}{2}\] in numerator and denominator, we get
\[{{\tan }^{-1}}\left( \dfrac{1-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{1+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}} \right)\]………………(9)
We know that, \[\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}\] ……………..(10)
Using equation (10), equation (9) can be written as
\[{{\tan }^{-1}}\left( \dfrac{1-\tan \dfrac{x}{2}}{1+\tan \dfrac{x}{2}} \right)\]
\[{{\tan }^{-1}}\left( \dfrac{1-\tan \dfrac{x}{2}}{1+1.\tan \dfrac{x}{2}} \right)\]……………(11)
We also know, \[\tan \dfrac{\pi }{4}=1\] ……………(12)
Using equation (12), equation (11) can be written as
\[{{\tan }^{-1}}\left( \dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}} \right)\]
\[={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)\]
\[=\dfrac{\pi }{4}-\dfrac{x}{2}\]
So, \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2}\] .
Therefore, LHS=RHS.
Hence, proved.

Note: In this question, we have to be careful during the simplification of the inverse tan function. As \[\tan \dfrac{\pi }{4}=1\] and also \[\tan \left( \dfrac{5\pi }{4} \right)=1\] . If we take \[\tan \left( \dfrac{5\pi }{4} \right)=1\] , then it will be wrong. As in RHS, we have \[\dfrac{\pi }{4}\] , so we have to continue with \[\tan \left( \dfrac{\pi }{4} \right)=1\] .