Animated Solution for Mathematics - Inverse Trigonometric Functions: If for some $\alpha ,\beta \in \mathbb{R}$ such that $\alpha \le \beta$, $\alpha +\beta =8$ and ${\sec }^2({\tan }^{-1}\alpha )+{\csc }^2({\cot }^{-1}\beta )=36,$ then ${\alpha }^2+\beta$ is equal to

Welcome, future engineer. Today, we are going to dismantle a problem that looks like a nightmare of trigonometry but is actually a beautiful, elegant dance of algebra. We are given $\alpha +\beta =8$ and the complex equation:

At first glance, this looks like a mess of inverse functions. But remember, in the world of JEE Advanced, complexity is often just a veil. Let us lift it.

We know that ${\tan }^{-1}(\alpha )$ is just an angle, let us call it $\theta$. So, $\tan (\theta )=\alpha$. The term ${\sec }^2({\tan }^{-1}\alpha )$ is simply ${\sec }^2(\theta )$. Using the fundamental identity $1+{\tan }^2(\theta )={\sec }^2(\theta )$, this becomes $1+{\alpha }^2$.

Similarly, for the second term, let $\varphi ={\cot }^{-1}(\beta )$, so $\cot (\varphi )=\beta$. The term ${\csc }^2(\varphi )$ becomes $1+{\cot }^2(\varphi )$, which is $1+{\beta }^2$. Just like that, the trigonometric nightmare has melted away into pure, clean algebra.

Now, substitute these back into our original equation:

Simplifying this, we get $2+{\alpha }^2+{\beta }^2=36$, which leads us to ${\alpha }^2+{\beta }^2=34$. We now have a system of two equations: $\alpha +\beta =8$ and ${\alpha }^2+{\beta }^2=34$.

To solve this system, we need the product $\alpha \beta$. We use the identity $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$. Plugging in our known values, we get:

This simplifies to $64=34+2\alpha \beta$. Solving for the product, we find $2\alpha \beta =30$, so $\alpha \beta =15$.

We have the sum $\alpha +\beta =8$ and the product $\alpha \beta =15$. This is a classic setup. Any two numbers with a known sum and product are the roots of the quadratic equation $t^2-(\text{sum})t+(\text{product})=0$.

So, our variables $\alpha$ and $\beta$ are the roots of the following equation:

Factoring this, we get $(t-3)(t-5)=0$. The roots are $3$ and $5$.

We are given the constraint $\alpha \le \beta$. This means $\alpha =3$ and $\beta =5$. Finally, we calculate the target expression:

We have arrived at the answer, 14. It was not about the trigonometry; it was about seeing the structure beneath the surface. Keep practicing this, and you will start seeing these patterns everywhere.