Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering only the principal values of the inverse trigonometric functions, the value of $\frac{3}{2}{\cos }^{-1}\sqrt{\frac{2}{2+{\pi }^2}}+\frac{1}{4}{\sin }^{-1}\frac{2\sqrt{2}\pi }{2+{\pi }^2}+{\tan }^{-1}\frac{\sqrt{2}}{\pi }$ is ______.

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are unraveling a tapestry.

When you first look at the expression

it feels like a chaotic jumble of constants and inverse functions. But take a deep breath; in the world of JEE Advanced, complexity is often just a mask for hidden symmetry.

Finding the Common Language

Our first goal is to bring all these disparate terms into a single, unified language. Let us look at the first term: $\theta ={\cos }^{-1}\sqrt{\frac{2}{2+{\pi }^2}}$.

If we visualize a right-angled triangle where the base is $\sqrt{2}$ and the hypotenuse is $\sqrt{2+{\pi }^2}$, the Pythagorean theorem reveals the perpendicular side to be:

Suddenly, the fog clears! We see that $\tan \theta =\frac{\pi }{\sqrt{2}}$, which means our first term is simply $\frac{3}{2}{\tan }^{-1}\frac{\pi }{\sqrt{2}}$.

The Trap of the Domain

Now, we turn our attention to the second term: ${\sin }^{-1}\frac{2\sqrt{2}\pi }{2+{\pi }^2}$. If you divide the numerator and denominator by 2, you get:

This is the classic form ${\sin }^{-1}\frac{2x}{1+x^2}$, where $x=\frac{\pi }{\sqrt{2}}$.

Here is where the trap lies. Many students blindly apply $2{\tan }^{-1}x$. But wait! Since $\pi \approx 3.14$ and $\sqrt{2}\approx 1.41$, our $x$ is approximately $2.22$.

Because $x>1$, we must use the identity ${\sin }^{-1}\frac{2x}{1+x^2}=\pi -2{\tan }^{-1}x$. This is the moment where the "JEE-level" thinking separates the crowd. We replace the second term with $\frac{1}{4}(\pi -2{\tan }^{-1}\frac{\pi }{\sqrt{2}})$.

The Elegant Collapse

Finally, we look at the third term: ${\tan }^{-1}\frac{\sqrt{2}}{\pi }$. This is just the reciprocal of our common variable.

Using the property ${\tan }^{-1}\frac{1}{y}={\cot }^{-1}y$, we transform this into ${\cot }^{-1}\frac{\pi }{\sqrt{2}}$. Now, let us assemble our pieces. The expression becomes:

Look at the coefficients of ${\tan }^{-1}\frac{\pi }{\sqrt{2}}$. We have $\frac{3}{2}-\frac{1}{2}=1$. The expression simplifies beautifully to:

We are left with the most beautiful identity in trigonometry: ${\tan }^{-1}y+{\cot }^{-1}y=\frac{\pi }{2}$.

Our expression collapses into:

Calculating the numerical value, we get approximately $2.356$, which rounds to $2.35$.

See? The chaos was merely an illusion. By staying calm, respecting the domain constraints, and looking for the underlying structure, we turned a terrifying problem into a moment of pure, mathematical harmony.