Let `S= sum_(r=1)^(5) cos (2r-1) (pi)/(11)` and `P=Pi_(r=1)^(4) cos (2^(r). (pi)/(15))`, then

Options

A. `log_(s)P=-4`
B. `P=3S`
C. `cosec S gt cosec P`
D. `tan^(-1)P lt tan^(-1)S`

Answer

To solve the problem, we need to compute the values of \( S \) and \( P \) as defined in the question.

### Step 1: Calculate \( S \)

The expression for \( S \) is given by:

\[ S = \sum_{r=1}^{5} \cos\left( (2r-1) \frac{\pi}{11} \right) \]

Calculating the individual terms:

- For \( r = 1 \): \( \cos\left( \frac{\pi}{11} \right) \)
- For \( r = 2 \): \( \cos\left( \frac{3\pi}{11} \right) \)
- For \( r = 3 \): \( \cos\left( \frac{5\pi}{11} \right) \)
- For \( r = 4 \): \( \cos\left( \frac{7\pi}{11} \right) \)
- For \( r = 5 \): \( \cos\left( \frac{9\pi}{11} \right) \)

Thus, we can write:

\[ S = \cos\left( \frac{\pi}{11} \right) + \cos\left( \frac{3\pi}{11} \right) + \cos\left( \frac{5\pi}{11} \right) + \cos\left( \frac{7\pi}{11} \right) + \cos\left( \frac{9\pi}{11} \right) \]

### Step 2: Use the Cosine Sum Formula

Using the identity for the sum of cosines, we can simplify the calculation. The sum of cosines can be expressed as:

\[ \sum_{k=0}^{n-1} \cos\left( a + kd \right) = \frac{\sin\left( \frac{nd}{2} \right) \cos\left( a + \frac{(n-1)d}{2} \right)}{\sin\left( \frac{d}{2} \right)} \]

For our case, \( a = \frac{\pi}{11} \), \( d = \frac{2\pi}{11} \), and \( n = 5 \):

\[ S = \frac{\sin\left( \frac{5 \cdot \frac{2\pi}{11}}{2} \right) \cos\left( \frac{\pi}{11} + \frac{(5-1) \cdot \frac{2\pi}{11}}{2} \right)}{\sin\left( \frac{\frac{2\pi}{11}}{2} \right)} \]

Calculating this gives:

\[ S = \frac{\sin\left( \frac{5\pi}{11} \right) \cos\left( \frac{9\pi}{11} \right)}{\sin\left( \frac{\pi}{11} \right)} \]

### Step 3: Calculate \( P \)

The expression for \( P \) is given by:

\[ P = \prod_{r=1}^{4} \cos\left( 2^r \frac{\pi}{15} \right) \]

Calculating the individual terms:

- For \( r = 1 \): \( \cos\left( \frac{2\pi}{15} \right) \)
- For \( r = 2 \): \( \cos\left( \frac{4\pi}{15} \right) \)
- For \( r = 3 \): \( \cos\left( \frac{8\pi}{15} \right) \)
- For \( r = 4 \): \( \cos\left( \frac{16\pi}{15} \right) \)

Thus, we can write:

\[ P = \cos\left( \frac{2\pi}{15} \right) \cdot \cos\left( \frac{4\pi}{15} \right) \cdot \cos\left( \frac{8\pi}{15} \right) \cdot \cos\left( \frac{16\pi}{15} \right) \]

### Step 4: Use Product-to-Sum Formulas

Using the product-to-sum identities, we can express \( P \) in terms of sine functions:

\[ P = \frac{1}{2^4} \cdot \frac{\sin\left( \frac{32\pi}{15} \right)}{\sin\left( \frac{2\pi}{15} \right)} \]

### Step 5: Compare \( S \) and \( P \)

From the calculations, we find:

- \( S = \frac{\sin\left( \frac{5\pi}{11} \right) \cos\left( \frac{9\pi}{11} \right)}{\sin\left( \frac{\pi}{11} \right)} \)
- \( P = \frac{1}{16} \cdot \frac{\sin\left( \frac{32\pi}{15} \right)}{\sin\left( \frac{2\pi}{15} \right)} \)

### Conclusion

By comparing \( S \) and \( P \), we conclude that:

\[ \tan^{-1}(S) > \tan^{-1}(P) \]

Thus, the correct option is \( \tan^{-1}(P) < \tan^{-1}(S) \).