If `a_(1), a_(2), …..,a_(n)` are in A.P. with common difference `d ne 0,` then the sum of the series sin `d[sec a_(1)sec a_(2) +..... sec a_(n-1) sec a_(n)]` is

Options

A. `cosec a_(n)-cosec a`
B. `cot a_(n)-cot a`
C. `sec a_(n)- sec a_(1)`
D. `tan a_(n)- tan a_(1)`

Explanation

To solve the problem, we need to find the sum of the series given by:

\[ S = \sin d \left( \sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \ldots + \sec a_{n-1} \sec a_n \right) \]

where \( a_1, a_2, \ldots, a_n \) are in arithmetic progression (A.P.) with common difference \( d \).

### Step 1: Express the terms in the series
Since \( a_k = a_1 + (k-1)d \) for \( k = 1, 2, \ldots, n \), we can express the secant terms as:

\[ \sec a_k = \sec(a_1 + (k-1)d) \]

### Step 2: Rewrite the series
The sum can be rewritten as:

\[ S = \sin d \left( \sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \ldots + \sec a_{n-1} \sec a_n \right) \]

### Step 3: Use the identity for secant
Recall that:

\[ \sec x = \frac{1}{\cos x} \]

Thus, we can rewrite the series in terms of cosine:

\[ S = \sin d \left( \frac{1}{\cos a_1 \cos a_2} + \frac{1}{\cos a_2 \cos a_3} + \ldots + \frac{1}{\cos a_{n-1} \cos a_n} \right) \]

### Step 4: Apply the sine difference identity
Using the identity \( \sin(a - b) = \sin a \cos b - \cos a \sin b \), we can express the differences:

\[ \sin(a_k - a_{k-1}) = \sin d \]

This allows us to express each term in the series as:

\[ \sin(a_k - a_{k-1}) = \sin d = \sin(a_k) \cos(a_{k-1}) - \cos(a_k) \sin(a_{k-1}) \]

### Step 5: Simplify the series
After substituting and simplifying, we find that the terms in the series will cancel out:

\[ S = \sin d \left( \tan a_2 - \tan a_1 + \tan a_3 - \tan a_2 + \ldots + \tan a_n - \tan a_{n-1} \right) \]

### Step 6: Recognize the telescoping nature
Notice that this series is telescoping:

\[ S = \sin d \left( \tan a_n - \tan a_1 \right) \]

### Final Result
Thus, the sum of the series is:

\[ S = \sin d \left( \tan a_n - \tan a_1 \right) \]