Animated Solution for Mathematics - Inverse Trigonometric Functions: $\cos ({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{33}{65})$ is equal to:

The problem asks us to evaluate the expression:

This expression appears intimidating, but the secret to conquering such problems in the JEE is the art of transformation. We will strip away the complexity by converting these inverse sine terms into a more manageable form.

Our first step is to demystify these inverse sine terms by treating each as an angle in a right-angled triangle. Let ${\theta }_1={\sin }^{-1}\frac{3}{5}$, which implies $\sin {\theta }_1=\frac{3}{5}$.

In a right triangle with opposite side $3$ and hypotenuse $5$, the base is $\sqrt{5^2-3^2}=4$. Thus, we can express this angle as:

Similarly, for ${\theta }_2={\sin }^{-1}\frac{5}{13}$, the opposite side is $5$ and the hypotenuse is $13$. The base is $\sqrt{13^2-5^2}=12$, leading to:

Now, we combine the first two angles using the identity ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$. Substituting $x=\frac{3}{4}$ and $y=\frac{5}{12}$:

The numerator is:

The denominator is:

Dividing the numerator by the denominator, we get:

Thus, the sum of the first two angles is ${\tan }^{-1}\frac{56}{33}$.

Now, consider the third term: ${\theta }_3={\sin }^{-1}\frac{33}{65}$. Here, the opposite side is $33$ and the hypotenuse is $65$. The base is $\sqrt{65^2-33^2}=\sqrt{4225-1089}=\sqrt{3136}=56$.

This gives us $\tan {\theta }_3=\frac{33}{56}$, which implies ${\theta }_3={\tan }^{-1}\frac{33}{56}$. Since $\tan {\theta }_3=\frac{33}{56}$, it follows that $\cot {\theta }_3=\frac{56}{33}$, or:

Our original expression now simplifies to:

Using the identity ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$, the argument inside the cosine becomes $\frac{\pi }{2}$. Therefore, the final result is: