`1/(cosalpha+cos3alpha)+1/(cosalpha+cos5alpha)....1/(cosalpha+cos(2n+1)alpha`

Options

A. `cosec alpha [tan(n+1)alpha-tan alpha]`
B. `sec alpha [ tan(n+1)alpha - tan alpha]`
C. `1/2 sec alpha[ tan (n+1)alpha - tan alpha]`
D. `1/2 cosec alpha[ tan (n+1) alpha - tan alpha]`

Answer

To solve the series

\[ S = \frac{1}{\cos \alpha + \cos 3\alpha} + \frac{1}{\cos \alpha + \cos 5\alpha} + \ldots + \frac{1}{\cos \alpha + \cos (2n + 1)\alpha} \]

we will use the trigonometric identity for the sum of cosines:

\[ \cos a + \cos b = 2 \cos \left(\frac{a+b}{2}\right) \cos \left(\frac{a-b}{2}\right) \]

### Step 1: Rewrite the first term

For the first term, we have:

\[ \cos \alpha + \cos 3\alpha = 2 \cos \left(\frac{\alpha + 3\alpha}{2}\right) \cos \left(\frac{\alpha - 3\alpha}{2}\right) = 2 \cos(2\alpha) \cos(-\alpha) = 2 \cos(2\alpha) \cos(\alpha) \]

Thus,

\[ \frac{1}{\cos \alpha + \cos 3\alpha} = \frac{1}{2 \cos(2\alpha) \cos(\alpha)} \]

### Step 2: Rewrite the second term

For the second term, we have:

\[ \cos \alpha + \cos 5\alpha = 2 \cos \left(\frac{\alpha + 5\alpha}{2}\right) \cos \left(\frac{\alpha - 5\alpha}{2}\right) = 2 \cos(3\alpha) \cos(-2\alpha) = 2 \cos(3\alpha) \cos(2\alpha) \]

Thus,

\[ \frac{1}{\cos \alpha + \cos 5\alpha} = \frac{1}{2 \cos(3\alpha) \cos(2\alpha)} \]

### Step 3: Generalize the term

Continuing this process, for the \(k\)-th term where \(k\) is odd, we can express:

\[ \cos \alpha + \cos (2k + 1)\alpha = 2 \cos \left(k\alpha + \alpha\right) \cos \left(k\alpha - \alpha\right) = 2 \cos((k+1)\alpha) \cos(k\alpha) \]

Thus, we can write:

\[ \frac{1}{\cos \alpha + \cos (2k + 1)\alpha} = \frac{1}{2 \cos((k+1)\alpha) \cos(k\alpha)} \]

### Step 4: Write the series

The series can now be expressed as:

\[ S = \sum_{k=1}^{n} \frac{1}{2 \cos((k+1)\alpha) \cos(k\alpha)} \]

### Step 5: Factor out constants

Factoring out the constant \( \frac{1}{2} \):

\[ S = \frac{1}{2} \sum_{k=1}^{n} \frac{1}{\cos((k+1)\alpha) \cos(k\alpha)} \]

### Step 6: Simplify the sum

Using the identity for tangent, we can express the sum in terms of tangent:

\[ \frac{1}{\cos((k+1)\alpha) \cos(k\alpha)} = \sec((k+1)\alpha) \sec(k\alpha) \]

This leads us to a telescoping series:

\[ S = \frac{1}{2} \left( \tan((n+1)\alpha) - \tan(\alpha) \right) \]

### Final Result

Thus, the final result for the series is:

\[ S = \frac{1}{2} \left( \tan((n+1)\alpha) - \tan(\alpha) \right) \]