If `a_(1), a_(2),……..a_(n)` are in A.P. with common difference d, then the sum of these series
`sin d ["cosec a"_(1) "cosec a"_(2) + "cosec a"_(2) "cosec a"_(3) + ......+ ....... + "cosec a"_(n-1) - "cosec a"_(n)]`

Options

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

Answer

To solve the problem, we need to evaluate the expression given in the question step by step.

### Step 1: Understanding the Arithmetic Progression (A.P.)
Given that \( a_1, a_2, \ldots, a_n \) are in A.P. with a common difference \( d \), we can express the terms as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- ...
- \( a_n = a + (n-1)d \)

### Step 2: Writing the Expression
We need to evaluate:
\[ \sin d \left[ \csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \ldots + \csc a_{n-1} \csc a_n \right] \]

### Step 3: Expanding the Terms
Using the definition of cosecant, we have:
\[ \csc a_i = \frac{1}{\sin a_i} \]
Thus, the expression becomes:
\[ \sin d \left[ \frac{1}{\sin a_1 \sin a_2} + \frac{1}{\sin a_2 \sin a_3} + \ldots + \frac{1}{\sin a_{n-1} \sin a_n} \right] \]

### Step 4: Applying the Sine Difference Formula
We can use the identity:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
to express the terms. Specifically, we can write:
\[ \sin d = \sin(a_2 - a_1) = \sin a_2 \cos a_1 - \cos a_2 \sin a_1 \]

### Step 5: Simplifying the Terms
Now, substituting back into our expression:
\[ \sin d \left[ \frac{\sin(a_2 - a_1)}{\sin a_1 \sin a_2} + \frac{\sin(a_3 - a_2)}{\sin a_2 \sin a_3} + \ldots \right] \]
This can be rewritten as:
\[ \sin d \left[ \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\sin a_1 \sin a_2} + \ldots \right] \]

### Step 6: Cancellation of Terms
Notice that in the series, many terms will cancel out:
- The \( \sin a_2 \) in the numerator of one term will cancel with the \( \sin a_2 \) in the denominator of the next term.

### Step 7: Final Result
After all cancellations, we will be left with:
\[ \cot a_1 - \cot a_n \]
Thus, the final expression simplifies to:
\[ \sin d \left( \cot a_1 - \cot a_n \right) \]

### Conclusion
The final result of the given expression is:
\[ \sin d \left( \cot a_1 - \cot a_n \right) \]