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 calculate the values of \( S \) and \( P \) and then compare \( \tan^{-1}(P) \) and \( \tan^{-1}(S) \).

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

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

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

Calculating each term in the summation:

- For \( r = 1 \): \( \cos\left(\frac{1 \pi}{11}\right) \)
- 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 have:

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

### Step 2: Simplify \( S \)

Using the property of cosine, we know that:

\[ \cos\left(\frac{9\pi}{11}\right) = -\cos\left(\frac{2\pi}{11}\right) \]
\[ \cos\left(\frac{7\pi}{11}\right) = -\cos\left(\frac{4\pi}{11}\right) \]

Thus, we can rewrite \( S \) as:

\[ S = \cos\left(\frac{\pi}{11}\right) + \cos\left(\frac{3\pi}{11}\right) - \cos\left(\frac{2\pi}{11}\right) - \cos\left(\frac{4\pi}{11}\right) \]

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

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

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

Calculating each term in the product:

- 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 have:

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

### Step 4: Simplify \( P \)

Using the property of cosine, we know that:

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

Thus, we can rewrite \( P \) as:

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

This simplifies to:

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

### Step 5: Compare \( \tan^{-1}(P) \) and \( \tan^{-1}(S) \)

Now we need to show that:

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

Since both \( S \) and \( P \) are positive, we can compare their values directly.

### Conclusion

After calculating \( S \) and \( P \), we can conclude that \( S > P \), which implies:

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

Thus, we have shown that \( 10^{-1}P < 10^{-1}S \).

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