Step 1: Choose a rationalisation approach:
Instead of converting to half angle squares directly, multiply the numerator and denominator of the fraction by $(\sqrt{1+\sin x}+\sqrt{1-\sin x})$.
This removes the difference of square roots from one part of the fraction and brings in $\cos x$ and $\sin x$ terms.
Step 2: Simplify the numerator after multiplying:
The new numerator is $(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2$.
Expanding this square gives $(1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}$.
\[ = 2 + 2\sqrt{1-\sin^2 x} = 2+2\cos x \]
Here $\cos x>0$ because $x\in(0,\pi/4)$ lies in the first quadrant, so the square root simplifies to $\cos x$ directly.
Step 3: Simplify the denominator after multiplying:
The new denominator is $(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2$, using the difference of squares formula $(p+q)(p-q)=p^2-q^2$ on the original fraction.
\[ = (1+\sin x)-(1-\sin x) = 2\sin x \]
So the whole fraction reduces to a simple expression in $\cos x$ and $\sin x$:
\[ \frac{2+2\cos x}{2\sin x} = \frac{1+\cos x}{\sin x} \]
Step 4: Use the standard half angle identity and invert:
Using the known identity $\dfrac{1+\cos x}{\sin x}=\cot\dfrac{x}{2}$ (obtained from $1+\cos x=2\cos^2\tfrac{x}{2}$ and $\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$), the fraction equals $\cot\dfrac{x}{2}$.
Since $x/2\in(0,\pi/8)$ lies inside the principal range $(0,\pi)$ of $\cot^{-1}$, we get $\cot^{-1}\left(\cot\tfrac{x}{2}\right)=\dfrac{x}{2}$ directly.
Final Answer:
This confirms the same result by a different route, so the identity is proved.
\[ \boxed{\cot^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2}} \]