Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f(x)=(\sin ({\tan }^{-1}x)+\sin ({\cot }^{-1}x){)}^2-1,|x|>1$. If $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}({\sin }^{-1}(f(x)))$ and $y(\sqrt{3})=\frac{\pi }{6}$, then $y(-\sqrt{3})$ is equal to:

Welcome, future engineers. Today, we are not just solving a calculus problem; we are embarking on a journey of simplification. Often, in the heat of a JEE Advanced exam, we see a function like $f(x)=(\sin ({\tan }^{-1}x)+\sin ({\cot }^{-1}x){)}^2-1$ and our instinct is to panic.

But I want you to take a deep breath and look at the structure. It is a composition of inverse trigonometric functions. Whenever you see ${\tan }^{-1}x$ or ${\cot }^{-1}x$, I want you to stop thinking about algebra and start thinking about geometry.

Imagine a right-angled triangle. If we define ${\tan }^{-1}x=\theta$, then $\tan \theta =\frac{x}{1}$. The opposite side is $x$, the adjacent is $1$, and the hypotenuse is $\sqrt{1+x^2}$.

Suddenly, $\sin ({\tan }^{-1}x)$ is not a scary function; it is just $\frac{x}{\sqrt{1+x^2}}$. By drawing these triangles, we strip away the complexity and reveal the simple, elegant ratios underneath.

Once we have our geometric intuition, the algebra begins to flow. We found that $\sin ({\tan }^{-1}x)=\frac{x}{\sqrt{1+x^2}}$ and $\sin ({\cot }^{-1}x)=\frac{1}{\sqrt{1+x^2}}$. When we add these together, we get $\frac{x+1}{\sqrt{1+x^2}}$.

Now, look at the function $f(x)$. We are squaring this sum and subtracting one. Expanding this, we get:

Watch the magic happen as we simplify:

The $x^2$ terms cancel, the $1$s cancel, and we are left with:

This is the "Aha!" moment. A complex expression has collapsed into a standard form. This is the beauty of mathematics—it rewards those who persist through the initial fog.

Now, we face the differential equation:

We substitute our simplified $f(x)$ to get $\frac{1}{2}\frac{d}{dx}({\sin }^{-1}(\frac{2x}{1+x^2}))$. Here, we must be vigilant. We apply the chain rule: the derivative of ${\sin }^{-1}(u)$ is $\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$.

This is where the domain constraint $|x|>1$ becomes critical. When we calculate $\sqrt{1-[f(x){]}^2}$, we are essentially calculating:

Because $|x|>1$, we know $x^2>1$, so $1-x^2$ is negative. We must write it as $x^2-1$. This is the kind of detail that separates the good from the great. By carefully handling this, the derivative simplifies beautifully to:

We have arrived at the home stretch. We have a separable differential equation: $dy=-\frac{1}{1+x^2}dx$. Integrating both sides is straightforward:

We use our boundary condition $y(\sqrt{3})=\frac{\pi }{6}$ to find that $c=\frac{\pi }{2}$. Our final function is $y=\frac{\pi }{2}-{\tan }^{-1}x$, which is the identity for ${\cot }^{-1}x$.

Finally, evaluating at $x=-\sqrt{3}$, we use the property ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$ to arrive at our answer:

You see? The complexity was just a mask. With patience, geometry, and rigorous calculus, the solution was always within your reach.