Animated Solution for Mathematics - Inverse Trigonometric Functions: ${\cot }^{-1}(\sqrt{\cos \alpha })-{\tan }^{-1}(\sqrt{\cos \alpha })=x$, then $\sin x=$

Welcome, future engineer. Today, we are not just solving an equation; we are choreographing a dance between inverse trigonometric functions.

When you look at the expression ${\cot }^{-1}(\sqrt{\cos \alpha })-{\tan }^{-1}(\sqrt{\cos \alpha })=x$, it might seem intimidating at first. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance waiting to be revealed.

Harmonizing the Functions

Our first step is to bring order to chaos. We have a ${\cot }^{-1}$ and a ${\tan }^{-1}$ function. Working with two different types of inverse functions is like trying to speak two languages at once.

Let's unify them. We know the fundamental identity: ${\cot }^{-1}y={\tan }^{-1}(\frac{1}{y})$.

By applying this to our first term, the expression ${\cot }^{-1}(\sqrt{\cos \alpha })$ transforms into ${\tan }^{-1}(\frac{1}{\sqrt{\cos \alpha }})$. Now, our equation is a beautiful, unified difference:

This is the harmony we were looking for.

The Power of the Difference Identity

Now that we have the form ${\tan }^{-1}A-{\tan }^{-1}B$, we can invoke the powerful difference identity:

Here, our $A$ is $\frac{1}{\sqrt{\cos \alpha }}$ and our $B$ is $\sqrt{\cos \alpha }$. When we substitute these into the formula, the expression inside the ${\tan }^{-1}$ becomes:

Look closely at the denominator—the $\sqrt{\cos \alpha }$ terms cancel out perfectly, leaving us with $1+1=2$. The numerator simplifies to $\frac{1-\cos \alpha }{\sqrt{\cos \alpha }}$.

Thus, we have:

The Geometric Bridge

We have successfully isolated $\tan x$. But the question asks for $\sin x$. How do we bridge the gap between tangent and sine? We build a triangle.

Imagine a right-angled triangle where the angle is $x$. By definition, $\tan x=\frac{\text{Perpendicular}}{\text{Base}}$.

So, we set the perpendicular $P=1-\cos \alpha$ and the base $B=2\sqrt{\cos \alpha }$. Now, we need the hypotenuse $H$. Using the Pythagorean theorem, $H=\sqrt{P^2+B^2}$.

As we calculated earlier, this simplifies beautifully to $H=1+\cos \alpha$. This is the moment where the math rewards your patience.

The Final Flourish

We are almost there. We know $\sin x=\frac{P}{H}=\frac{1-\cos \alpha }{1+\cos \alpha }$.

While this is a correct answer, it does not match our options. This is where we use the half-angle identities, the secret weapon of every JEE topper.

We know that $1-\cos \alpha =2{\sin }^2(\frac{\alpha }{2})$ and $1+\cos \alpha =2{\cos }^2(\frac{\alpha }{2})$. Substituting these in, we get:

The twos cancel, and we are left with ${\tan }^2(\frac{\alpha }{2})$.

And there it is. The complexity has dissolved, leaving behind a simple, elegant result. You didn't just solve a problem; you navigated a logical path. Keep this mindset, and no problem will ever be too difficult for you.