If $\theta_1, \theta_2, \theta_3, \ldots, \theta_n$ are in A.P., whose common difference is $d$, show that $\sec\theta_1 \sec\theta_2 + \sec\theta_2 \sec\theta_3 + \ldots + \sec\theta_{n-1} \sec\theta_n = \dfrac{\tan\theta_n - \tan\theta_1}{\sin d}$.

If θ₁, θ₂, θ₃, …, θₙ are in A.P., whose common difference is d, show that secθ₁ secθ₂ + secθ₂ secθ₃ + … + secθ

NCERT Exemplar · Q14

\theta_1, \theta_2, \theta_3, \ldots, \theta_n

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Concept understanding — Sum Of Products

Sum of Products — The Intuition First

Imagine you're buying a fruit basket. The shop has a rule: you must pick exactly one fruit from each of several groups. Group A has apples and bananas. Group B has oranges and mangoes. How many different baskets can you make?

You'd list them: (apple, orange), (apple, mango), (banana, orange), (banana, mango). That's $2 \times 2 = 4$ baskets.

Now, suppose each basket has a price that depends on which fruits you pick. The total money you'd spend if you bought every possible basket exactly once is the sum of the prices of all those baskets.

That's the core idea of Sum of Products: when you have a situation where you choose one item from each of several independent categories, the total "value" across all possible combinations is the sum of the products of the individual values.

The Precise Statement

\sum_{i_1=1}^{n_1} \sum_{i_2=1}^{n_2} \cdots \sum_{i_k=1}^{n_k} \left( a_{i_1}^{(1)} \cdot a_{i_2}^{(2)} \cdots a_{i_k}^{(k)} \right) = \left( \sum_{i_1=1}^{n_1} a_{i_1}^{(1)} \right) \left( \sum_{i_2=1}^{n_2} a_{i_2}^{(2)} \right) \cdots \left( \sum_{i_k=1}^{n_k} a_{i_k}^{(k)} \right)

In plain English: The sum over all combinations of products equals the product of the sums.

Let that sink in. It's not obvious — it's a beautiful distributive property that works because multiplication distributes over addition.

Why It Works — A Simple Example

Take two small groups:

Group 1: numbers $a_1, a_2$
Group 2: numbers $b_1, b_2$

All possible products: $a_1b_1 + a_1b_2 + a_2b_1 + a_2b_2$

Factor it: $a_1(b_1 + b_2) + a_2(b_1 + b_2) = (a_1 + a_2)(b_1 + b_2)$

That's it. The sum of all products is just the product of the sums. This extends to any number of groups.

Note

This is not the same as "product of sums" (which is a different expression). The order matters: Sum of Products = Product of Sums, but Product of Sums ≠ Sum of Products in general.

Where You'll Meet It

Probability: If you roll two dice, the sum of probabilities of all outcomes is $(\frac16 + \frac16 + \cdots)(\frac16 + \cdots) = 1 \times 1 = 1$ .

Combinatorics: Counting total number of combinations — each group has $n_i$ choices, so total combinations = $n_1 \times n_2 \times \cdots \times n_k$ . That's a special case where each "value" is just 1.

Algebra: Expanding $(x+2)(x+3)$ gives $x^2 + 5x + 6$ — that's a sum of products (each term is a product of one term from each bracket).

Tip

When you see a problem that says "find the sum of all possible products formed by taking one element from each set", immediately think: product of the sums. It saves enormous calculation.

A Common Mistake …

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The sum telescopes when each term $\sec\theta_k \sec\theta_{k+1}$ is rewritten using the identity $\tan(A) - \tan(B) = \sin(A-B)\sec A \sec B$ , leading to the result $\dfrac{\tan\theta_n - \tan\theta_1}{\sin d}$ .

The core idea here is telescoping — a technique where a sum collapses because successive terms cancel. But the given terms don't look like they cancel directly. The trick is to express each product $\sec\theta_k \sec\theta_{k+1}$ as a difference of tangents.

Why tangents? Because the derivative of $\tan x$ is $\sec^2 x$ , and the difference formula for tangent involves secants. Since $\theta_1, \theta_2, \dots$ are in arithmetic progression, the difference between consecutive angles is constant ( $d$ ), which makes the sine of that difference a constant factor — perfect for pulling out of the sum.

Let’s build this step by step.

Recall the tangent difference identity For any two angles $A$ and $B$ ,

\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B} = \sin(A-B) \sec A \sec B.

This is the bridge: it turns a product of secants into a difference of tangents divided by a sine.

Apply it to consecutive terms in the AP Since $\theta_{k+1} - \theta_k = d$ , we have

\tan\theta_{k+1} - \tan\theta_k = \sin d \cdot \sec\theta_k \sec\theta_{k+1}.

Therefore,

\sec\theta_k \sec\theta_{k+1} = \frac{\tan\theta_{k+1} - \tan\theta_k}{\sin d}.

Tip

This single step is the entire secret. Once you see it, the sum becomes a simple telescoping series.

Write the sum using this expression The given sum $S$ is $S = \sum_{k=1}^{n-1} \sec\theta_k \sec\theta_{k+1} = \frac{1}{\sin d} \sum_{k=1}^{n-1} \left( \tan\theta_{k+1} - \tan\theta_k \right).$ …

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