Animated Solution for Mathematics - Trigonometry: Let $S=\left\{\theta \in \left(0,\frac{\pi }{2}\right):{\sum }_{m=1}^9\sec \left(\theta +(m-1)\frac{\pi }{6}\right)\sec \left(\theta +\frac{m\pi }{6}\right)=-\frac{8}{\sqrt{3}}\right\}$. Then

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of secant functions.

When you see a summation like this in a JEE Advanced paper, your heart might skip a beat. But remember, in the world of competitive mathematics, complexity is often just a mask for elegance. Let us peel back that mask together.

First, let us look at the general term of our summation, which we will call $T_m$. The expression is:

It looks intimidating, doesn't it? But let us simplify our lives. Let us define the first angle as $\alpha =\theta +(m-1)\frac{\pi }{6}$ and the second as $\beta =\theta +m\frac{\pi }{6}$.

Now, look at the difference between these two angles:

This is our master key. The difference is a constant!

Working with secants in a sum is a nightmare. Let us convert them to cosines:

Now, here is the trick that separates the masters from the novices. We want to turn this product in the denominator into a difference in the numerator. We multiply and divide by $\sin (\beta -\alpha )$.

Since $\beta -\alpha =\frac{\pi }{6}$, we are essentially multiplying by $\sin (\frac{\pi }{6})$, which is $\frac{1}{2}$.

So, the expression becomes:

Using the compound angle formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$, the numerator becomes $\sin \beta \cos \alpha -\cos \beta \sin \alpha$. When we divide this by $\cos \alpha \cos \beta$, the expression simplifies beautifully to:

Now, substitute $\alpha$ and $\beta$ back in. We have:

When we sum this from $m=1$ to $9$, we get a telescoping series. Imagine a chain of falling dominoes!

The second term of the first bracket cancels the first term of the second bracket, and so on. Only the very first and very last terms survive:

We simplify $\frac{9\pi }{6}$ to $\frac{3\pi }{2}$. We know that $\tan (\frac{3\pi }{2}+\theta )=-\cot \theta$.

Thus, our sum becomes $-2(\cot \theta +\tan \theta )$. Equating this to the given value $-\frac{8}{\sqrt{3}}$, we get:

Converting to sine and cosine, we get $\frac{1}{\sin \theta \cos \theta }=\frac{4}{\sqrt{3}}$, which leads us to:

Given $\theta \in (0,\frac{\pi }{2})$, we have $2\theta \in (0,\pi )$. The solutions are $2\theta =\frac{\pi }{3}$ and $2\theta =\frac{2\pi }{3}$.

This gives us $\theta =\frac{\pi }{6}$ and $\theta =\frac{\pi }{3}$. The sum of these values is $\frac{\pi }{2}$. You have conquered the monster!