If `x_(1),x_(2),x_(3),...,x_(n)` are in A.P. whose common difference is `alpha,` then the value of `sinalpha(secx_(1)secx_(2)+secx_(2)secx_(3)+...+secx_(n-1)secx_(n))`is

Options

A. `(sin(n-1)alpha)/(cosx_(1)cosx_(n))`
B. `(sinn alpha)/(cosx_(1)cosx_(n))`
C. `sin(n-1)alphacosx_(1)cosx_(n)`
D. `sinn alphacosx_(1)cosx_(n)`

Answer

To solve the problem, we need to find the value of

\[ \sin \alpha \left( \sec x_1 \sec x_2 + \sec x_2 \sec x_3 + \ldots + \sec x_{n-1} \sec x_n \right) \]

where \( x_1, x_2, \ldots, x_n \) are in arithmetic progression (A.P.) with a common difference \( \alpha \).

### Step 1: Express the terms in the sum using cosine

Recall that

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

Thus, we can rewrite the expression as:

\[ \sin \alpha \left( \frac{1}{\cos x_1 \cos x_2} + \frac{1}{\cos x_2 \cos x_3} + \ldots + \frac{1}{\cos x_{n-1} \cos x_n} \right) \]

### Step 2: Factor out \(\sin \alpha\)

This gives us:

\[ \sin \alpha \left( \frac{1}{\cos x_1 \cos x_2} + \frac{1}{\cos x_2 \cos x_3} + \ldots + \frac{1}{\cos x_{n-1} \cos x_n} \right) \]

### Step 3: Use the sine difference identity

We know that the common difference \( \alpha \) can be expressed in terms of the angles. For two angles \( x_i \) and \( x_{i+1} \):

\[ \sin(x_{i+1} - x_i) = \sin \alpha = \sin(x_i + \alpha - x_i) = \sin x_i \cos \alpha - \cos x_i \sin \alpha \]

### Step 4: Apply the sine difference formula

Using the sine difference formula, we can express each term in the sum as:

\[ \sin(x_{i+1}) \cos x_i - \cos(x_{i+1}) \sin x_i \]

### Step 5: Sum the series

Now, summing these terms:

\[ \sum_{i=1}^{n-1} \left( \sin(x_{i+1}) \cos x_i - \cos(x_{i+1}) \sin x_i \right) \]

This results in a telescoping series, where most terms will cancel out.

### Step 6: Simplify the result

After cancellation, we are left with:

\[ \sin x_n - \sin x_1 \]

### Step 7: Final expression

Thus, we can express our original expression as:

\[ \sin \alpha \cdot \frac{\sin(x_n) - \sin(x_1)}{\cos x_1 \cos x_n} \]

### Step 8: Substitute \( x_n \)

Since \( x_n = x_1 + (n-1)\alpha \), we can substitute this into our expression:

\[ \sin \alpha \cdot \frac{\sin(x_1 + (n-1)\alpha) - \sin x_1}{\cos x_1 \cos(x_1 + (n-1)\alpha)} \]

### Step 9: Final simplification

Using the sine difference identity:

\[ \sin(x_n - x_1) = \sin((n-1)\alpha) \]

Thus, we have:

\[ \sin(n-1)\alpha \cdot \frac{1}{\cos x_1 \cos x_n} \]

### Conclusion

The final result is:

\[ \frac{\sin(n-1)\alpha}{\cos x_1 \cos x_n} \]

### Final Answer

The value of

\[ \sin \alpha \left( \sec x_1 \sec x_2 + \sec x_2 \sec x_3 + \ldots + \sec x_{n-1} \sec x_n \right) = \frac{\sin(n-1)\alpha}{\cos x_1 \cos x_n} \]