Animated Solution for Mathematics - Differentiation: If $y(\alpha )=\sqrt{2\left(\frac{\tan \alpha +\cot \alpha }{1+{\tan }^2\alpha }\right)+\frac{1}{{\sin }^2\alpha }}$, $\alpha \in \left(\frac{3\pi }{4},\pi \right)$ then $\frac{dy}{d\alpha }$ at $\alpha =\frac{5\pi }{6}$ is

Welcome, future engineers. Today, we are going to dissect a problem that looks like a nightmare but is actually a beautifully orchestrated dance of trigonometric identities.

When you first look at the expression

your instinct might be to panic. It is messy, it has a square root, and it involves multiple trigonometric functions. But remember, in the world of JEE Advanced, complexity is often a mask for elegance. Let us peel back the layers together.

The Simplification Strategy

Our first mission is to simplify the fraction inside the square root: $\frac{\tan \alpha +\cot \alpha }{1+{\tan }^2\alpha }$. The denominator, $1+{\tan }^2\alpha$, is a classic identity that screams to be replaced by ${\sec }^2\alpha$.

Now, look at the numerator: $\tan \alpha +\cot \alpha$. If we convert these into sines and cosines, we get:

Now, let's combine these pieces. We have $\frac{1}{\sin \alpha \cos \alpha }÷{\sec }^2\alpha$. Since ${\sec }^2\alpha =\frac{1}{{\cos }^2\alpha }$, dividing by ${\sec }^2\alpha$ is the same as multiplying by ${\cos }^2\alpha$.

The expression becomes:

Just like that, the entire fraction has vanished, replaced by the elegant $\cot \alpha$.

The Hidden Perfect Square

Now, let's substitute this back into our original function. The expression inside the square root becomes $2\cot \alpha +\frac{1}{{\sin }^2\alpha }$. We know that $\frac{1}{{\sin }^2\alpha }={\csc }^2\alpha$.

So, we have:

Here is where the magic happens. We know the identity ${\csc }^2\alpha =1+{\cot }^2\alpha$. Substituting this in, we get:

Does this look familiar? It is the algebraic expansion of $(a+b{)}^2$, where $a=\cot \alpha$ and $b=1$. Thus, $y(\alpha )=\sqrt{(1+\cot \alpha {)}^2}$.

The Modulus Trap

This is the moment where many students lose marks. We have $\sqrt{(1+\cot \alpha {)}^2}$. The common mistake is to write $1+\cot \alpha$. But we must remember the fundamental rule: $\sqrt{x^2}=|x|$.

So, $y(\alpha )=|1+\cot \alpha |$.

We are given the interval $\alpha \in (\frac{3\pi }{4},\pi )$. Let's analyze the behavior of $\cot \alpha$ in this region. At $\alpha =\frac{3\pi }{4}$, $\cot \alpha =-1$. As $\alpha$ approaches $\pi$, $\cot \alpha$ decreases toward $-\infty$.

Therefore, in this interval, $\cot \alpha <-1$. This implies that $1+\cot \alpha <0$. Since the expression inside the modulus is negative, we must remove the modulus bars by multiplying the expression by $-1$:

The Final Derivative

Now, the problem is trivial. We need to find $\frac{dy}{d\alpha }$ for $y(\alpha )=-1-\cot \alpha$. Differentiating with respect to $\alpha$, the derivative of $-1$ is $0$, and the derivative of $-\cot \alpha$ is $-(-{\csc }^2\alpha )={\csc }^2\alpha$.

Finally, we evaluate this at $\alpha =\frac{5\pi }{6}$. We know that $\sin (\frac{5\pi }{6})=\sin (\pi -\frac{\pi }{6})=\sin (\frac{\pi }{6})=\frac{1}{2}$.

Therefore, $\csc (\frac{5\pi }{6})=2$. Squaring this, we get:

And there you have it! Through careful simplification and a keen eye for the modulus, we have tamed the beast. The final answer is 4. Keep practicing this level of rigor, and you will find that even the most intimidating problems are just puzzles waiting to be solved.