Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $M$ and $m$ respectively be the maximum and minimum values of the function $f(x)={\tan }^{-1}(\sin x+\cos x)$ in $[0,\frac{\pi }{2}]$, Then the value of $\tan (M-m)$ is equal to:

We are given the function $f(x)={\tan }^{-1}(\sin x+\cos x)$ defined on the interval $x\in [0,\frac{\pi }{2}]$. Our objective is to determine the maximum value $M$ and the minimum value $m$ of this function, and subsequently evaluate $\tan (M-m)$.

Since the function ${\tan }^{-1}(u)$ is a strictly increasing function, the extrema of $f(x)$ correspond directly to the extrema of the inner function $g(x)=\sin x+\cos x$. We can focus our analysis entirely on $g(x)$ and apply the inverse tangent at the final stage.

Whenever an expression takes the form $a\sin x+b\cos x$, we utilize the harmonic identity:

Setting $a=1$ and $b=1$, we find $\sqrt{a^2+b^2}=\sqrt{2}$. Thus, the function simplifies to:

Given the domain $x\in [0,\frac{\pi }{2}]$, we shift the interval by adding $\frac{\pi }{4}$ to all components. This yields the new domain for the argument of the sine function:

We now evaluate the range of $\sin (x+\frac{\pi }{4})$ within the interval $[\frac{\pi }{4},\frac{3\pi }{4}]$. The sine function increases from $\frac{1}{\sqrt{2}}$ at $\frac{\pi }{4}$ to a maximum of $1$ at $\frac{\pi }{2}$, then decreases back to $\frac{1}{\sqrt{2}}$ at $\frac{3\pi }{4}$.

Consequently, the range of $g(x)$ is:

Applying the outer function ${\tan }^{-1}(u)$, we identify the extrema:

We are tasked with evaluating $\tan (M-m)=\tan ({\tan }^{-1}(\sqrt{2})-\frac{\pi }{4})$. We apply the trigonometric subtraction formula:

Substituting $A={\tan }^{-1}(\sqrt{2})$ and $B=\frac{\pi }{4}$, we have $\tan A=\sqrt{2}$ and $\tan B=1$. The expression becomes:

To rationalize the denominator, we multiply the numerator and denominator by $(\sqrt{2}-1)$:

The final result is: