Animated Solution for Mathematics - Inverse Trigonometric Functions: Find the value of : $\cos (2{\cos }^{-1}x+{\sin }^{-1}x)$ at $x=\frac{1}{5}$, where $0\le {\cos }^{-1}x\le \pi$ and $-\pi /2\le {\sin }^{-1}x\le \pi /2$.

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that, at first glance, might seem like a tangled mess of inverse functions.

We are tasked with finding the value of $\cos (2{\cos }^{-1}x+{\sin }^{-1}x)$ at $x=\frac{1}{5}$.

In the world of JEE, complexity is often just a mask for elegance waiting to be revealed. Let us peel back that mask together.

The core of our challenge lies in the argument of the cosine function: $2{\cos }^{-1}x+{\sin }^{-1}x$. We do not have a direct identity for this specific combination.

However, mathematics is the art of seeing patterns. Let us rewrite the argument as:

Why do this? Because now, we have exposed the hidden gem: ${\cos }^{-1}x+{\sin }^{-1}x$.

We know that for any $x\in [-1,1]$, the sum is exactly $\frac{\pi }{2}$. By making this simple split, we have transformed our expression into:

Now, we are looking at $\cos \left(\frac{\pi }{2}+{\cos }^{-1}x\right)$. Let us treat ${\cos }^{-1}x$ as a single angle, $\theta$.

Our expression is now $\cos \left(\frac{\pi }{2}+\theta \right)$. This brings us to the allied angle formulas.

In the second quadrant, the cosine function is negative, and the function itself co-functions into sine. Thus:

Substituting our $\theta$ back, we get $-\sin ({\cos }^{-1}x)$. Do not forget that negative sign; it is a common trap.

We are now left with the task of evaluating $-\sin ({\cos }^{-1}x)$. Let $\theta ={\cos }^{-1}x$, which implies $\cos \theta =x$.

Since $x=\frac{1}{5}$ is positive, our angle $\theta$ must lie in the first quadrant. Imagine a right-angled triangle where the base is $x$ and the hypotenuse is $1$.

Using the Pythagorean theorem, the perpendicular side is $\sqrt{1^2-x^2}=\sqrt{1-x^2}$. Therefore:

Our expression, $-\sin ({\cos }^{-1}x)$, now becomes $-\sqrt{1-x^2}$.

We have done the heavy lifting. Now, we simply substitute $x=\frac{1}{5}$ into our simplified expression:

Squaring the fraction gives us $\frac{1}{25}$. So, we have:

Simplifying the square root, we get $-\frac{\sqrt{24}}{5}$. Since $\sqrt{24}=2\sqrt{6}$, our final result is:

We started with a daunting inverse trigonometric expression and, through strategic splitting, identity application, and geometric visualization, we arrived at a precise, elegant answer.