1. Interpret the function

The given function is $f(x)=\tan^{-1}(\sin x+\cos x), \qquad x\in\left[0,\frac{\pi}{2}\right].$

We need the maximum value $M$ and minimum value $m$ of $f(x)$ on this interval, and then compute $\tan(M-m).$

2. Find the range of $\sin x+\cos x$ on $\left[0,\frac{\pi}{2}\right]$

Use the identity $\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right).$

Now as $x\in\left[0,\frac{\pi}{2}\right],$ we get $x+\frac{\pi}{4}\in\left[\frac{\pi}{4},\frac{3\pi}{4}\right].$

On this interval, $\sin\left(x+\frac{\pi}{4}\right)$ has:

• maximum value $1$ at $x=\frac{\pi}{4}$,
• minimum value $\frac{\sqrt{2}}{2}$ at the endpoints $x=0,\frac{\pi}{2}$.

Therefore, $\sin x+\cos x\in[1,\sqrt{2}].$

So:

• minimum of $\sin x+\cos x$ is $1$,
• maximum of $\sin x+\cos x$ is $\sqrt{2}$.

3. Use monotonicity of $\tan^{-1}x$

Since $\tan^{-1}x$ is an increasing function, the extrema of $f(x)=\tan^{-1}(\sin x+\cos x)$ occur at the extrema of $\sin x+\cos x$.

Hence, $m=\tan^{-1}(1)=\frac{\pi}{4},$ $M=\tan^{-1}(\sqrt{2}).$

Thus, $M-m=\tan^{-1}(\sqrt{2})-\frac{\pi}{4}.$

We need $\tan(M-m)=\tan\left(\tan^{-1}(\sqrt{2})-\frac{\pi}{4}\right).$

4. Apply tangent subtraction formula

Let $A=\tan^{-1}(\sqrt{2}), \qquad B=\frac{\pi}{4}.$ Then $\tan A=\sqrt{2}, \qquad \tan B=1.$

Using $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$ we get $\tan(M-m)=\frac{\sqrt{2}-1}{1+\sqrt{2}}.$

Now simplify: