Animated Solution for Mathematics - Inverse Trigonometric Functions: If $\alpha ={\cos }^{-1}(\frac{3}{5}),\beta ={\tan }^{-1}(\frac{1}{3})$, where $0<\alpha ,\beta <\frac{\pi }{2}$, then $\alpha -\beta$ is equal to :

Decoding the Hidden Angles

Welcome, future engineer. Today, we are going to dismantle a problem that often intimidates students, not because it is difficult, but because it looks like a foreign language.

When you see expressions like $\alpha ={\cos }^{-1}(\frac{3}{5})$ and $\beta ={\tan }^{-1}(\frac{1}{3})$, your first instinct might be to reach for a calculator or panic about the lack of 'standard' angles. But here is the secret: you don't need to know what $\alpha$ or $\beta$ are individually. You only need to know how they behave.

The Geometric Translation

Think of inverse trigonometric functions as labels for right-angled triangles. When we say $\alpha ={\cos }^{-1}(\frac{3}{5})$, we are simply saying, 'Imagine a right-angled triangle where the adjacent side is $3$ and the hypotenuse is $5$.'

Before we do anything else, let's translate these into a common language. Tangent is the most powerful tool in our arsenal for combining angles.

For $\alpha$, if the base is $3$ and the hypotenuse is $5$, the perpendicular must be:

Thus, $\tan \alpha =\frac{4}{3}$.

Now, look at $\beta$. It is already given as ${\tan }^{-1}(\frac{1}{3})$. This is a gift! It tells us directly that $\tan \beta =\frac{1}{3}$.

We have successfully translated both angles into the language of tangents. We are no longer dealing with abstract inverse functions; we are dealing with simple ratios.

The Bridge of Compound Angles

We need to find $\alpha -\beta$. We have the tangents of the individual angles, so we reach for the compound angle identity:

This formula is the bridge. It allows us to perform arithmetic on the angles themselves by performing algebra on their ratios. Let's substitute our values:

Take a moment to appreciate the elegance here. The numerator is $\frac{4}{3}-\frac{1}{3}=\frac{3}{3}=1$. The denominator is $1+\frac{4}{9}=\frac{13}{9}$.

When we divide these, we get:

The Final Reconstruction

We have found that $\tan (\alpha -\beta )=\frac{9}{13}$. However, if you look at the options provided, they are expressed in terms of ${\sin }^{-1}$. Do not be discouraged! This is not a dead end; it is a simple conversion.

We treat $(\alpha -\beta )$ as a single angle in a new right-angled triangle. If the tangent of this angle is $\frac{9}{13}$, then the opposite side is $9$ and the adjacent side is $13$.

To find the sine, we need the hypotenuse. Using the Pythagorean theorem:

We can simplify $\sqrt{250}$ as $\sqrt{25\times 10}=5\sqrt{10}$. Now, the sine of our angle is simply the ratio of the opposite side to the hypotenuse:

Therefore, the final result is:

Look at what we just accomplished. We didn't need to know the exact value of $\alpha$ (which is roughly $53.13^{\circ }$) or $\beta$ (roughly $18.43^{\circ }$). We navigated the problem using the structural properties of trigonometry.

This is the essence of JEE Advanced physics and mathematics: it is rarely about brute-force calculation and almost always about choosing the right representation. You translated the problem, built a bridge, and reconstructed the final answer. That is the mindset of a topper.