If $$f'(x) = \tan^{-1}(\sec x + \tan x)$$, $$-\frac{\pi}{2} < x < \frac{\pi}{2}$$ and $$f(0) = 0$$, then $$f(1)$$ is equal to:

We have the derivative $$f'(x)=\tan^{-1}(\sec x+\tan x)$$ defined for $$-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$$ and the initial condition $$f(0)=0$$. Our task is to compute $$f(1)$$.

First we try to simplify the inside of the inverse-tangent. A standard trigonometric identity is

$$\sec x+\tan x=\dfrac{1+\sin x}{\cos x}.$$

Another well-known identity connects the half-angle substitution $$t=\tan\dfrac{x}{2}$$ with the usual sine and cosine: we have

$$\sin x=\dfrac{2t}{1+t^{2}}, \qquad \cos x=\dfrac{1-t^{2}}{1+t^{2}}.$$

Substituting these into $$\dfrac{1+\sin x}{\cos x}$$ gives

$$\dfrac{1+\sin x}{\cos x} =\dfrac{1+\dfrac{2t}{1+t^{2}}}{\dfrac{1-t^{2}}{1+t^{2}}} =\dfrac{(1+t^{2})+2t}{1-t^{2}} =\dfrac{(t+1)^{2}}{(1-t)(1+t)} =\dfrac{t+1}{1-t}.$$

But the tangent‐addition formula states

$$\tan\!\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr) =\dfrac{1+\tan\tfrac{x}{2}}{1-\tan\tfrac{x}{2}} =\dfrac{t+1}{1-t}.$$

Hence we have proved the compact identity

$$\sec x+\tan x=\tan\!\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr).$$

Using this result in the derivative, we obtain

$$f'(x)=\tan^{-1}\!\Bigl(\tan\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr)\Bigr).$$

Now observe that for the given domain $$-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$$ we have $$-\dfrac{\pi}{4}<\dfrac{x}{2}<\dfrac{\pi}{4},$$ so

$$0<\tfrac{\pi}{4}+\tfrac{x}{2}<\dfrac{\pi}{2}.$$

The angle $$\tfrac{\pi}{4}+\tfrac{x}{2}$$ therefore lies inside the principal branch $$\bigl(-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigr)$$ of the inverse-tangent, and in this interval the functions $$\tan$$ and $$\tan^{-1}$$ cancel each other exactly. Hence

$$f'(x)=\tfrac{\pi}{4}+\tfrac{x}{2}.$$

To find $$f(x)$$ we integrate term by term:

$$f(x)=\int f'(x)\,dx =\int\!\Bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\Bigr)dx =\tfrac{\pi}{4}\,x+\dfrac{x^{2}}{4}+C,$$ where $$C$$ is the constant of integration.

The condition $$f(0)=0$$ gives

$$0=f(0)=\tfrac{\pi}{4}\cdot 0+\dfrac{0^{2}}{4}+C \;\Longrightarrow\;C=0.$$ So the function itself is

$$f(x)=\tfrac{\pi}{4}\,x+\dfrac{x^{2}}{4}.$$

Finally, evaluating at $$x=1$$ yields

$$f(1)=\tfrac{\pi}{4}\cdot 1+\dfrac{1^{2}}{4} =\dfrac{\pi+1}{4}.$$

Hence, the correct answer is Option A.