Animated Solution for Mathematics - Inverse Trigonometric Functions: If $y(x)={\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$, $x\in \left(\frac{\pi }{2},\pi \right)$, then $\frac{dy}{dx}$ at $x=\frac{5\pi }{6}$ is:

Taming the Inverse Trigonometric Expression

My dear student, I know exactly what you are thinking when you look at this problem. You see a massive, intimidating fraction buried inside a cotangent inverse, and your first instinct might be to reach for the chain rule and start a long, painful differentiation process.

But stop! Take a breath. In the world of JEE Advanced, whenever you encounter a complex inverse trigonometric function, the goal is rarely to differentiate it directly. The goal is to see the hidden beauty, the algebraic structure waiting to be revealed.

Let us embark on a journey to simplify this expression, step by step.

The Algebraic Surgery

Look at the expression inside the cotangent inverse:

It looks like a monster, but it is built on a very simple foundation. We need to eliminate those square roots by creating perfect squares inside them.

Recall the fundamental identity $1={\sin }^2(x/2)+{\cos }^2(x/2)$ and the double-angle formula $\sin x=2\sin (x/2)\cos (x/2)$. When we substitute these into our terms, something magical happens:

Suddenly, the square roots are no longer terrifying. They are just waiting to be cancelled.

The Absolute Value Trap

Here is where the elite students separate themselves from the crowd. When you cancel a square root with a square, you must write the result as an absolute value: $\sqrt{X^2}=|X|$.

So, our expression becomes $|\sin (x/2)+\cos (x/2)|$ and $|\sin (x/2)-\cos (x/2)|$. We are given $x\in (\pi /2,\pi )$, which means $x/2\in (\pi /4,\pi /2)$.

In this interval, $\sin (x/2)$ is always positive and greater than $\cos (x/2)$. Therefore, $\sin (x/2)+\cos (x/2)$ is positive, and $\sin (x/2)-\cos (x/2)$ is also positive.

This allows us to open the absolute value signs without any sign changes. The expression simplifies to:

The Elegant Cancellation

Watch as the complexity vanishes. In the numerator, the $\cos (x/2)$ terms cancel out, leaving $2\sin (x/2)$. In the denominator, the $\sin (x/2)$ terms cancel out, leaving $2\cos (x/2)$.

Our massive fraction has collapsed into:

Now, our function is simply $y(x)={\cot }^{-1}(\tan (x/2))$. To cancel the ${\cot }^{-1}$, we need a cotangent inside.

Using the complementary angle property, $\tan (x/2)=\cot (\pi /2-x/2)$. Thus, $y(x)={\cot }^{-1}(\cot (\pi /2-x/2))$.

Since our angle $(\pi /2-x/2)$ lies within the principal branch of the cotangent inverse, they cancel perfectly, leaving us with $y(x)=\pi /2-x/2$.

The Final Victory

Now, the differentiation is trivial. The derivative of a constant $\pi /2$ is $0$, and the derivative of $-x/2$ is $-1/2$.

The derivative is:

This is a constant, independent of $x$. Whether $x=5\pi /6$ or any other value in the domain, the slope remains $-1/2$. You have conquered the monster!