Animated Solution for Mathematics - Inverse Trigonometric Functions: The greater of the two angles $A=2{\tan }^{-1}(2\sqrt{2}-1)$ and $B=3{\sin }^{-1}(1/3)+{\sin }^{-1}(3/5)$ is .........

A Mathematical Duel

Welcome, fellow traveler of the JEE journey. Today, we face a classic problem that tests not just your ability to compute, but your ability to reason.

We are presented with two formidable angles:

At first glance, they look like a mess of radicals and inverse functions. But remember: in mathematics, when the direct path is blocked, we don't force our way through—we find a vantage point.

The Strategic Pivot

Directly calculating these values is a trap. Instead, let's find a 'reference point' to act as a judge.

We suspect that $\frac{2\pi }{3}$ is the perfect pivot. If we can prove that $A$ is greater than this value and $B$ is less than it, the comparison becomes trivial. It is like placing two objects on a scale and seeing which one tips the balance against a known weight.

Unmasking Angle A

Let's look at $A=2{\tan }^{-1}(2\sqrt{2}-1)$. The argument $2\sqrt{2}-1$ is the key.

We know $\sqrt{2}\approx 1.414$, so $2(1.414)-1=2.828-1=1.828$. Now, compare this to $\sqrt{3}\approx 1.732$.

Since $1.828>1.732$, we can confidently state that $2\sqrt{2}-1>\sqrt{3}$. Because the function ${\tan }^{-1}(x)$ is strictly increasing, we have:

Multiplying by $2$, we arrive at the beautiful conclusion:

Taming the Beast of Angle B

Now, for $B=3{\sin }^{-1}(\frac{1}{3})+{\sin }^{-1}(\frac{3}{5})$. That $3{\sin }^{-1}(\frac{1}{3})$ looks intimidating, but we have the triple angle identity in our arsenal:

Substituting $x=\frac{1}{3}$, we get:

Now, $B={\sin }^{-1}(\frac{23}{27})+{\sin }^{-1}(\frac{3}{5})$. Let's test these against our pivot, $\frac{\pi }{3}$. We know $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}\approx 0.866$.

For the first term, $\frac{23}{27}\approx 0.852$. Since $0.852<0.866$, it follows that ${\sin }^{-1}(\frac{23}{27})<\frac{\pi }{3}$.

For the second term, $\frac{3}{5}=0.6$. Since $0.6<0.866$, it follows that ${\sin }^{-1}(0.6)<\frac{\pi }{3}$.

Adding these two inequalities together, we find that:

We have successfully trapped our values. We know $A>\frac{2\pi }{3}$ and $B<\frac{2\pi }{3}$.

The logic is now undeniable: $A$ must be the greater angle. This problem wasn't about brute-force calculation; it was about understanding the behavior of functions and using inequalities to navigate the landscape.