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 a common difference \( d \).

### Step 1: Express the terms in A.P.
Since \( a_1, a_2, \ldots, a_n \) are in A.P., we can express them as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- ...
- \( a_n = a + (n-1)d \)

### Step 2: Write the secant terms
The secant terms can be expressed as:
- \( \sec a_1 = \sec a \)
- \( \sec a_2 = \sec(a + d) \)
- \( \sec a_3 = \sec(a + 2d) \)
- ...
- \( \sec a_n = \sec(a + (n-1)d) \)

### Step 3: Rewrite the sum
The sum can be rewritten as:
\[ S = \sin d \left( \sec a \sec(a + d) + \sec(a + d) \sec(a + 2d) + \ldots + \sec(a + (n-2)d) \sec(a + (n-1)d) \right) \]

### Step 4: Use the identity for secant
Using the identity \( \sec x \sec y = \frac{1}{\cos x \cos y} \), we can express the terms:
\[ \sec a_i \sec a_{i+1} = \frac{1}{\cos a_i \cos a_{i+1}} \]

### Step 5: Apply the sine difference identity
We can apply the sine difference identity:
\[ \sin(a + d) - \sin a = 2 \cos\left(a + \frac{d}{2}\right) \sin\left(\frac{d}{2}\right) \]
to express the terms in a more manageable form.

### Step 6: Simplify the expression
Using the telescoping nature of the series:
\[ S = \sin d \left( \frac{\sin(a + d) - \sin a}{\cos a \cos(a + d)} + \frac{\sin(a + 2d) - \sin(a + d)}{\cos(a + d) \cos(a + 2d)} + \ldots \right) \]
This will simplify to:
\[ S = \sin d \left( \tan(a + (n-1)d) - \tan a \right) \]

### Final Result
Thus, the final result for the sum of the series is:
\[ S = \tan(a + (n-1)d) - \tan a \]