Animated Solution for Mathematics - Inverse Trigonometric Functions: The sum of the absolute maximum and absolute minimum values of the function $f(x)={\tan }^{-1}(\sin x-\cos x)$ in the interval $[0,\pi ]$ is

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are choreographing a dance between trigonometric functions and their inverse counterparts.

We are looking at the function $f(x)={\tan }^{-1}(\sin x-\cos x)$ on the closed interval $[0,\pi ]$. It looks intimidating, but every complex problem is just a series of simple, elegant steps waiting to be uncovered.

Before we can tackle the ${\tan }^{-1}$ function, we must understand the heart of the expression: $g(x)=\sin x-\cos x$. This is a classic harmonic form.

Whenever you see $\sin x$ and $\cos x$ together, your intuition should scream 'Harmonic Addition Theorem!' By multiplying and dividing by $\sqrt{1^2+(-1{)}^2}=\sqrt{2}$, we transform the expression into:

Recognizing that $\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$ and $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, we invoke the identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$. Suddenly, the chaos settles into the beautiful, rhythmic form:

Now, we must respect the boundaries. We are restricted to $x\in [0,\pi ]$.

If we shift our perspective by subtracting $\frac{\pi }{4}$, our new domain for the argument $\theta =x-\frac{\pi }{4}$ becomes $[-\frac{\pi }{4},\frac{3\pi }{4}]$. Imagine walking along the unit circle; you start at $-\frac{\pi }{4}$ (the fourth quadrant) and sweep all the way to $\frac{3\pi }{4}$ (the second quadrant).

In this journey, the sine function starts at $\sin (-\frac{\pi }{4})=-\frac{1}{\sqrt{2}}$, climbs up to its maximum of $1$ at $\theta =\frac{\pi }{2}$, and then descends to $\sin (\frac{3\pi }{4})=\frac{1}{\sqrt{2}}$.

Multiplying by our amplitude $\sqrt{2}$, we find the range of $g(x)$ to be $[-1,\sqrt{2}]$.

We are almost there. We have $f(x)={\tan }^{-1}(g(x))$. Because the inverse tangent function is strictly increasing, it preserves the order of its inputs.

The absolute minimum of $f(x)$ occurs at the minimum of $g(x)$, and the absolute maximum occurs at the maximum of $g(x)$.

For the minimum: $f_{min}={\tan }^{-1}(-1)=-\frac{\pi }{4}$.

For the maximum: $f_{max}={\tan }^{-1}(\sqrt{2})$.

To match the elegance of our options, we convert ${\tan }^{-1}(\sqrt{2})$ into a ${\cos }^{-1}$ form. If $\tan \theta =\frac{\sqrt{2}}{1}$, we visualize a right triangle with opposite side $\sqrt{2}$ and adjacent side $1$.

By the Pythagorean theorem, the hypotenuse is $\sqrt{1^2+(\sqrt{2}{)}^2}=\sqrt{3}$. Thus, $\cos \theta =\frac{1}{\sqrt{3}}$, or $\theta ={\cos }^{-1}(\frac{1}{\sqrt{3}})$.

The sum of our absolute maximum and minimum is simply:

Look at that result. It is not just a number; it is the culmination of geometric insight, domain analysis, and the beautiful properties of monotonic functions. You didn't just solve a problem; you mastered the logic behind it.