Statement I `Pi_(r=1)^(n) (1+sec 2^(r)theta)=tan 2^(n)theta cot theta`
Statement II `Pi_(r=1)^(n) cos (2^(r-1)theta)=(sin(2^(n)theta))/(2^(n)sin theta)`

Options

A. A
B. B
C. C
D. D

Answer

To solve the given statements, we will analyze each statement step by step.

### Statement I:
\[ \prod_{r=1}^{n} (1 + \sec(2^r \theta)) = \tan(2^n \theta) \cot(\theta) \]

**Step 1: Rewrite the left-hand side.**
We start with the product:
\[ \prod_{r=1}^{n} (1 + \sec(2^r \theta)) \]
Using the identity \(\sec x = \frac{1}{\cos x}\), we can rewrite this as:
\[ \prod_{r=1}^{n} \left(1 + \frac{1}{\cos(2^r \theta)}\right) = \prod_{r=1}^{n} \left(\frac{\cos(2^r \theta) + 1}{\cos(2^r \theta)}\right) \]

**Step 2: Simplify the product.**
This can be expressed as:
\[ \frac{\prod_{r=1}^{n} (\cos(2^r \theta) + 1)}{\prod_{r=1}^{n} \cos(2^r \theta)} \]

**Step 3: Analyze the right-hand side.**
The right-hand side is:
\[ \tan(2^n \theta) \cot(\theta) = \frac{\sin(2^n \theta)}{\cos(2^n \theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} = \frac{\sin(2^n \theta) \cos(\theta)}{\sin(\theta) \cos(2^n \theta)} \]

**Step 4: Establish the equality.**
We need to show that:
\[ \frac{\prod_{r=1}^{n} (\cos(2^r \theta) + 1)}{\prod_{r=1}^{n} \cos(2^r \theta)} = \frac{\sin(2^n \theta) \cos(\theta)}{\sin(\theta) \cos(2^n \theta)} \]

Using the double angle identities and properties of sine and cosine, we can verify that both sides are equal, confirming that Statement I is true.

### Statement II:
\[ \prod_{r=1}^{n} \cos(2^{r-1} \theta) = \frac{\sin(2^n \theta)}{2^n \sin(\theta)} \]

**Step 1: Rewrite the left-hand side.**
The left-hand side is:
\[ \prod_{r=1}^{n} \cos(2^{r-1} \theta) \]
This expands to:
\[ \cos(\theta) \cos(2\theta) \cos(4\theta) \ldots \cos(2^{n-1}\theta) \]

**Step 2: Use the product-to-sum identities.**
There is a known identity for the product of cosines:
\[ \prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin(\theta)} \]
This directly gives us the right-hand side.

**Step 3: Establish the equality.**
Thus, we can conclude that:
\[ \prod_{r=1}^{n} \cos(2^{r-1} \theta) = \frac{\sin(2^n \theta)}{2^n \sin(\theta)} \]
which confirms that Statement II is also true.

### Conclusion:
Both Statement I and Statement II are true.