Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of ${\sin }^{-1}(\frac{12}{13})-{\sin }^{-1}(\frac{3}{5})$ is equal to :

A Journey into Inverse Trigonometry

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are peeling back the layers of a trigonometric puzzle.

When you look at an expression like ${\sin }^{-1}(\frac{12}{13})-{\sin }^{-1}(\frac{3}{5})$, your brain might immediately reach for a calculator. But in the world of JEE Advanced, we don't need calculators—we have the elegance of geometry and the power of identities.

Defining the Players

Let us start by grounding ourselves. When we write $\alpha ={\sin }^{-1}(\frac{12}{13})$, we are simply saying that $\alpha$ is an angle whose sine is $\frac{12}{13}$.

Imagine a right-angled triangle where the opposite side is $12$ and the hypotenuse is $13$. By the Pythagorean theorem, the adjacent side must be $\sqrt{13^2-12^2}=5$.

Similarly, for $\beta ={\sin }^{-1}(\frac{3}{5})$, we have a triangle with opposite side $3$ and hypotenuse $5$, making the adjacent side $4$. We are looking for the difference between these two angles, $\alpha -\beta$.

The Power of Identities

Instead of hunting for the values of $\alpha$ and $\beta$, we use the subtraction identity for inverse sine functions:

This formula is not just a random string of symbols; it is the physical manifestation of the sine subtraction identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$.

When we substitute $x=\frac{12}{13}$ and $y=\frac{3}{5}$, we are essentially calculating the sine of the difference of two angles. The term $\sqrt{1-y^2}$ is simply $\cos ({\sin }^{-1}y)$, and $\sqrt{1-x^2}$ is $\cos ({\sin }^{-1}x)$. It is all just geometry!

The Calculation

Let us perform the arithmetic with care. We calculate the square roots:

Now, substitute these back into our master equation:

We have arrived at ${\sin }^{-1}(\frac{33}{65})$. This is where the true test of a JEE aspirant begins: the ability to transform your result into the required form.

The Final Transformation

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$. If we let $\theta ={\sin }^{-1}(\frac{33}{65})$, then $\sin \theta =\frac{33}{65}$.

To find $\cos \theta$, we use the identity $\cos \theta =\sqrt{1-{\sin }^2\theta }$:

Thus, $\theta ={\cos }^{-1}(\frac{56}{65})$. Using our complementary identity, we can rewrite this as:

And there it is! The elegance of the result reveals itself. By manipulating the identity, we have matched our result to the required form.

Remember, in mathematics, the journey is just as important as the destination. You have successfully navigated through identities, geometry, and algebraic manipulation to reach the final answer.