Animated Solution for Mathematics - Trigonometry: If $u=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$ then the difference between the maximum and minimum values of $u^2$ is given by

Unraveling the Expression

Have you ever looked at an expression that seemed designed to intimidate you? The expression $u=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$ is exactly that.

It looks like a tangled mess of square roots and trigonometric functions. But in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask together.

The Power of Squaring

When you see square roots, your first instinct should be to eliminate them. We are looking for the range of $u^2$, so let's square the entire expression.

Using the identity $(x+y{)}^2=x^2+y^2+2xy$, we treat the first square root as $x$ and the second as $y$. This gives us:

The Elegant Simplification

Look at the first part of our expansion: $(a^2{\cos }^2\theta +b^2{\sin }^2\theta )+(a^2{\sin }^2\theta +b^2{\cos }^2\theta )$. If we group the $a^2$ terms and the $b^2$ terms, we get $a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\sin }^2\theta +{\cos }^2\theta )$.

Since ${\sin }^2\theta +{\cos }^2\theta =1$, this massive chunk simplifies to just $a^2+b^2$. It is a moment of pure mathematical relief!

Taming the Cross-Product

Now, we face the cross-product term: $2\sqrt{XY}$. Expanding the product $XY$ inside the square root is where the real work happens.

After careful multiplication, we get:

Grouping the terms with common factors, we have $XY=(a^4+b^4){\sin }^2\theta {\cos }^2\theta +a^2{b}^2({\sin }^4\theta +{\cos }^4\theta )$.

The Trigonometric Transformation

To simplify this, we use double-angle identities. Recall that ${\sin }^2\theta {\cos }^2\theta =\frac{1}{4}{\sin }^{2}2\theta$ and ${\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{\sin }^{2}2\theta$.

Substituting these into our expression for $XY$ leads us to:

This is the turning point. Our expression for $u^2$ now becomes:

Finding the Extremes

Now, the variation of $u^2$ depends entirely on ${\sin }^{2}2\theta$, which ranges from $0$ to $1$.

- Minimum: When ${\sin }^{2}2\theta =0$, $u_{\text{min}}^2=a^2+b^2+2ab=(a+b{)}^2$.

- Maximum: When ${\sin }^{2}2\theta =1$, $u_{\text{max}}^2=(a^2+b^2)+2\sqrt{a^2{b}^2+\frac{(a^2-b^2{)}^2}{4}}=2(a^2+b^2)$.

Finally, the difference between the maximum and minimum values is $2(a^2+b^2)-(a+b{)}^2=a^2+b^2-2ab=(a-b{)}^2$.

We have arrived at the solution, and it is as clean as it is beautiful. Keep practicing, and you will find that even the most daunting problems have a logical, elegant path to the answer.