Animated Solution for Mathematics - Inverse Trigonometric Functions: $2\pi -\left({\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}\right)$ is equal to:

Welcome, future engineer! Today, we are going to dismantle a problem that often intimidates students at first glance. We are looking at the expression:

It looks like a mess of inverse functions, but I want you to see it differently. Think of each ${\sin }^{-1}$ term not as a complex function, but simply as an angle. Let us call the sum of these angles $A$. Our goal is to find $2\pi -A$.

Why Tangent is Our Best Friend

Working with ${\sin }^{-1}$ addition formulas is like trying to solve a puzzle with missing pieces—it is messy and inefficient. Instead, we use a powerful tool in our JEE arsenal: the conversion to ${\tan }^{-1}$.

For any $\sin \theta =\frac{p}{h}$, we can construct a right-angled triangle where the perpendicular is $p$ and the hypotenuse is $h$. By the Pythagorean theorem, the base is $b=\sqrt{h^2-p^2}$. Thus, $\tan \theta =\frac{p}{b}$.

The Triangle Conversions

For the first term, ${\sin }^{-1}\frac{4}{5}$, we have a perpendicular of $4$ and a hypotenuse of $5$. The base is $\sqrt{5^2-4^2}=3$. So, ${\sin }^{-1}\frac{4}{5}={\tan }^{-1}\frac{4}{3}$.

Moving to the second term, ${\sin }^{-1}\frac{5}{13}$, we have a perpendicular of $5$ and a hypotenuse of $13$. The base is $\sqrt{13^2-5^2}=12$. Thus, ${\sin }^{-1}\frac{5}{13}={\tan }^{-1}\frac{5}{12}$.

Finally, for ${\sin }^{-1}\frac{16}{65}$, the perpendicular is $16$ and the hypotenuse is $65$. The base is $\sqrt{65^2-16^2}=63$. So, ${\sin }^{-1}\frac{16}{65}={\tan }^{-1}\frac{16}{63}$. Our expression $A$ is now:

The Algebra of Tangents

Now, we combine the first two terms using the identity ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$. First, check the product:

Since $\frac{5}{9}<1$, we can proceed directly. The numerator is $\frac{4}{3}+\frac{5}{12}=\frac{16+5}{12}=\frac{21}{12}$. The denominator is $1-\frac{20}{36}=\frac{16}{36}$.

Dividing these gives:

So, the sum of the first two terms is ${\tan }^{-1}\frac{63}{16}$.

The Elegant Collapse

Look at what we have now:

Do you see the beauty here? The arguments are exact reciprocals! We know the identity ${\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{2}$ for $x>0$.

Therefore, $A=\frac{\pi }{2}$. Finally, our original objective was $2\pi -A$. Substituting our result, we get:

And there you have it—a complex-looking problem reduced to a simple, elegant conclusion. Keep practicing this conversion technique; it is a lifesaver in the exam hall! The final answer is $\frac{3\pi }{2}$.