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Q: Why can't we just expand the cosecants using the addition formula? That seems more direct.

You could, but that leads to a nightmare of algebraic expansion. In JEE Advanced, the goal isn't just to solve, but to solve elegantly. Telescoping is the intended path because it turns a complex sum into a simple subtraction of two terms. Always look for the 'telescoping' pattern when you see products in the denominator.

Q: How did you know the difference was $\frac{\pi}{4}$?

Look at the arguments: $A = \theta + \frac{(m-1)\pi}{4}$ and $B = \theta + \frac{m\pi}{4}$. Subtracting them is a standard check. Always look for the constant difference in these types of series. If the difference is constant, you can almost always use the cotangent identity.

Q: Why do we get two values for $\theta$?

Because $\sin(2\theta) = \frac{1}{2}$ has two solutions in the range $(0, \pi)$. Since $0 < \theta < \frac{\pi}{2}$, we have $0 < 2\theta < \pi$. Both solutions are valid within the given domain, so we must include both.

Question

For $0 < 	heta < \pi/2$, the solution(s) of $\sum_{m=1}^6 \csc (	heta + \frac{(m-1)\pi}{4}) \csc (	heta + \frac{m\pi}{4}) = 4\sqrt{2}$ is (are)

Options:

(A) $\pi/4$

(B) $\pi/6$

(C) $\pi/12$

(D) $5\pi/12$

Sigma Percentile · Accepted Answer

### The Art of the Telescoping Series

Welcome, future engineer. Today, we are not just solving a trigonometric equation; we are uncovering a hidden rhythm.

When you first look at the expression

$$\sum_{m=1}^6 \csc \left(	heta + \frac{(m-1)\pi}{4}ight) \csc \left(	heta + \frac{m\pi}{4}ight) = 4\sqrt{2}$$

it is natural to feel a sense of intimidation. It looks like a wall of cosecants, but in the world of JEE Advanced, intimidation is just a signal that you are about to learn something beautiful.

### Phase 1: The Cosecant Conundrum

Cosecants are rarely our friends in algebraic manipulation. They are the reciprocals of sines, and sines are where the real magic happens.

Let us rewrite our general term $T_m$ as:

$$T_m = \frac{1}{\sin\left(	heta + \frac{(m-1)\pi}{4}ight) \sin\left(	heta + \frac{m\pi}{4}ight)}$$

Now, we have a product of sines in the denominator. This is the classic setup for a telescoping series. We need to turn this product into a difference using the identity:

$$\frac{\sin(B-A)}{\sin A \sin B} = \cot A - \cot B$$

This is our most powerful tool here. It allows us to decompose a fraction into a difference of two terms, which is exactly what we need to make the intermediate terms vanish.

### Phase 2: The Grand Cancellation

Let us define our angles. Let $A = 	heta + \frac{(m-1)\pi}{4}$ and $B = 	heta + \frac{m\pi}{4}$.

When we calculate the difference $B - A$, the $	heta$ terms cancel out, leaving us with a constant difference of $\frac{\pi}{4}$. This is the key!

To use our identity, we need $\sin(B-A)$ in the numerator. Currently, our numerator is just $1$. We can fix this by multiplying and dividing by $\sin(\frac{\pi}{4})$.

Since $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, we are effectively multiplying the entire sum by $\sqrt{2}$. Now, our general term becomes:

$$T_m = \sqrt{2} \left[ \cot\left(	heta + \frac{(m-1)\pi}{4}ight) - \cot\left(	heta + \frac{m\pi}{4}ight) ight]$$

### Phase 3: The Final Act

Now, watch the magic happen. As we sum from $m=1$ to $6$, the terms expand:

$$\sqrt{2} \left[ \left(\cot 	heta - \cot\left(	heta + \frac{\pi}{4}ight)ight) + \dots + \left(\cot\left(	heta + \frac{5\pi}{4}ight) - \cot\left(	heta + \frac{6\pi}{4}ight)ight) ight]$$

Notice how the second part of the first term cancels the first part of the second term? This domino effect continues until only the very first and very last terms remain.

We are left with:

$$\sqrt{2} \left[ \cot 	heta - \cot\left(	heta + \frac{3\pi}{2}ight) ight]$$

Using the allied angle formula, $\cot(	heta + \frac{3\pi}{2}) = -	an 	heta$, so our sum simplifies to $\sqrt{2}(\cot 	heta + 	an 	heta)$.

Equating this to $4\sqrt{2}$, we find $\cot 	heta + 	an 	heta = 4$. This simplifies to:

$$\frac{1}{\sin 	heta \cos 	heta} = 4 \Rightarrow \sin 2	heta = \frac{1}{2}$$

Solving for $	heta$ in the range $(0, \frac{\pi}{2})$, we get **$	heta = \frac{\pi}{12}$** and **$	heta = \frac{5\pi}{12}$**.

You have just mastered the art of the telescoping series. Keep this logic in your toolkit; it is a weapon that will serve you well in the exam hall.

Animated Solution for Mathematics - Trigonometry: For $0 < θ < π /2$ , the solution(s) of $\sum_{m = 1}^{6} csc (θ + \frac{( m - 1 ) π}{4}) csc (θ + \frac{mπ}{4}) = 42$ is (are)

Select Answer:

Visualized Solution

Analyzing the Trigonometric Sum

Converting Cosecants to Sines

The Difference Identity Tool

Finding the Constant Difference

Constructing the Telescoping Term

Simplifying to Cotangent Differences

Expanding and Canceling Terms

Evaluating the Surviving Terms

Solving for $sin 2 θ$

Finding the Final Solutions

The Sigma Insight: Trigonometric Ratios and Identities

The Art of the Telescoping Series

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For 0<θ<π/2, the solution(s) of ∑m=16​csc(θ+4(m−1)π​)csc(θ+4mπ​)=42​ is (are)