The value of the expression ${{\tan }^{-1}}\left( \sec x+\tan x \right)$ in the interval $\left( 0,\dfrac{\pi }{2} \right)$ is
[a] $\dfrac{\pi }{4}-x$
[b] $x-\dfrac{\pi }{4}$
[c] $x+\dfrac{3\pi }{4}$
[d] $\dfrac{x}{2}+\dfrac{\pi }{4}$

Hint: Use the fact that $\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$ and hence prove that $\tan x+\sec x=\dfrac{1+\sin x}{\cos x}$. Use the fact that $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$ and $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ and hence prove that $\tan x+\sec x=\dfrac{1+\cos \left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)}$. Put $\dfrac{\pi }{2}-x=t$ and use the fact that $1+\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$. Hence prove that $\sec x+\tan x=\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)$. Use the fact that ${{\tan }^{-1}}\left( \tan x \right)=x\forall x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and hence find the value of ${{\tan }^{-1}}\left( \sec x+\tan x \right),x\in \left( 0,\dfrac{\pi }{2} \right)$

Complete step by step answer:
Let $S=\sec x+\tan x$
We know that $\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$. Using these identities, we get
$S=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}$
Taking cosx as LCM, we get
$S=\dfrac{1+\sin x}{\cos x}$
We know that $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$ and $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$. Using these identities, we get
$S=\dfrac{1+\cos \left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)}$
Put $\dfrac{\pi }{2}-x=t$, we get
$S=\dfrac{1+\cos t}{\sin t}$
We know that $1+\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$.
Using the above identity, we get
$S=\dfrac{2{{\cos }^{2}}\dfrac{t}{2}}{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}$
Cancelling the common factor $2\cos \dfrac{t}{2}$ from the numerator and denominator, we get
$S=\dfrac{\cos \dfrac{t}{2}}{\sin \dfrac{t}{2}}$
We know that $\dfrac{\cos x}{\sin x}=\cot x$
Using the above identity, we get
$S=\cot \dfrac{t}{2}$
We know that $\cot x=\tan \left( \dfrac{\pi }{2}-x \right)$
Using the above identity, we get
$S=\tan \left( \dfrac{\pi }{2}-\dfrac{t}{2} \right)$
Reverting to original variable, we get
$S=\tan \left( \dfrac{\pi }{2}-\dfrac{\dfrac{\pi }{2}-x}{2} \right)$
We know that $\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}$
Hence, we have
$S=\tan \left( \dfrac{\pi }{2}-\dfrac{\pi }{4}+\dfrac{x}{2} \right)=\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)$
Hence, we have
$\sec x+\tan x=\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)$
Hence, we have
${{\tan }^{-1}}\left( \sec x+\tan x \right)={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right) \right)$
We have $x\in \left( 0,\dfrac{\pi }{2} \right)$
Hence, we have $\dfrac{x}{2}\in \left( 0,\dfrac{\pi }{4} \right)\Rightarrow \dfrac{x}{2}+\dfrac{\pi }{4}\in \left( \dfrac{\pi }{4},\dfrac{\pi }{2} \right)$
We know that ${{\tan }^{-1}}\left( \tan x \right)=x\forall x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$
Using the above results, we have
${{\tan }^{-1}}\left( \sec x+\tan x \right)=\dfrac{\pi }{4}+\dfrac{x}{2}\forall x\in \left( 0,\dfrac{\pi }{2} \right)$

So, the correct answer is “Option D”.

Note: [1] A general way to find the values of ${{\tan }^{-1}}\tan x,{{\sin }^{-1}}\sin x$ etc is to put these function equal to y and hence arrive at equations of the form $\tan x=\tan y,\sin x=\sin y$ etc.
Finally use the fact that $\tan y=\tan x\Rightarrow y=n\pi +x\in \mathbb{Z}$ where n is suitable chose so that the value of y is in the interval $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ the principal branch of tanx. This is a general method and the student should use this method when he forgets the definition of ${{\tan }^{-1}}\tan x$ as
\[{{\tan }^{-1}}\tan x=\left\{ \begin{matrix}
   \vdots \\
   \pi +x,x\in \left( -\dfrac{3\pi }{2},\dfrac{-\pi }{2} \right) \\
   x,x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right) \\
   \pi -x,x\in \left( \dfrac{\pi }{2},\dfrac{3\pi }{2} \right) \\
   \vdots \\
\end{matrix} \right.\]