Animated Solution for Mathematics - Differentiation: Let $f(\theta )=\sin \left({\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\cos 2\theta }}\right)\right)$, where $-\frac{\pi }{4}<\theta <\frac{\pi }{4}$. Then the value of $\frac{d}{d(\tan \theta )}(f(\theta ))$ is

Welcome, students. Today we tackle a problem that looks like a monster, but it is actually a paper tiger. When you first see $f(\theta )=\sin ({\tan }^{-1}(\frac{\sin \theta }{\sqrt{\cos 2\theta }}))$, it is natural to feel a spike of anxiety.

The nested functions, the square root, and the inverse tangent all scream complexity. But in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back the layers together.

The first step in any high-level problem is to stop looking at the whole and start looking at the parts. We are given the domain $-\frac{\pi }{4}<\theta <\frac{\pi }{4}$.

This is not just a formality; it is a hint. It ensures that $\cos 2\theta$ is positive, keeping our square root real. Now, let us simplify our life.

Let $\alpha ={\tan }^{-1}(\frac{\sin \theta }{\sqrt{\cos 2\theta }})$. This transforms our function into the much friendlier $f(\theta )=\sin \alpha$. We have effectively reduced a mountain to a molehill.

Whenever you see a trigonometric ratio like $\tan \alpha =\frac{\sin \theta }{\sqrt{\cos 2\theta }}$, you should immediately visualize a right-angled triangle. Let the angle be $\alpha$.

The opposite side is $\sin \theta$, and the adjacent side is $\sqrt{\cos 2\theta }$. To find $\sin \alpha$, we need the hypotenuse $H$. By the Pythagorean theorem:

This is where the magic happens.

Expanding the squares, we get $H^2={\sin }^2\theta +\cos 2\theta$. Now, recall the double angle identity: $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$.

Substituting this into our equation for $H^2$, we get:

The ${\sin }^2\theta$ terms cancel out perfectly, leaving us with $H^2={\cos }^2\theta$. Thus, $H=\cos \theta$ (since $\cos \theta >0$ in our domain).

Now, we return to our function $f(\theta )=\sin \alpha$. From our triangle:

Our massive, scary function has collapsed into simple $\tan \theta$. The question asks for $\frac{d}{d(\tan \theta )}(f(\theta ))$.

Since $f(\theta )=\tan \theta$, we are simply differentiating $\tan \theta$ with respect to $\tan \theta$. The result is 1.

See? The monster was just a shadow. Keep practicing, keep visualizing, and you will master these challenges.