1. Given function

We need the range of

We must choose the interval $B$ such that $f:(-1,1)\to B$ is both one-one and onto.

2. Use the tangent double-angle identity

Recall:

Let

Since $x\in(-1,1)$, we have

Then

So,

Now since

and $\tan^{-1}(\tan y)=y$ for $y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, we get

3. Determine the range

Since $x\in(-1,1)$ and $\tan^{-1}x$ is strictly increasing,

Therefore,

Also, the endpoints are not attained because $x=-1$ and $x=1$ are not in the domain.

Hence the range is

4. Check one-one

The function

is strictly increasing on $(-1,1)$ because $\tan^{-1}x$ is strictly increasing. Hence $f$ is one-one.

5. Check onto for the options

For $f:(-1,1)\to B$ to be onto, $B$ must equal the range of $f$, i.e.

Now evaluate options:

• A: $\left(0,\frac{\pi}{2}\right)$ — not correct, misses negative values.
• B: $\left[0,\frac{\pi}{2}\right)$ — not correct, misses negative values and includes $0$ as left endpoint formatting issue only, but still not full range.
• C: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ — not correct, includes endpoints not attained.
• D: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ — correct.

6. Final answer

Therefore, the interval $B$ for which $f$ is both one-one and onto is

So the correct option is D.