The sum `1/(sin4 5^(@)sin4 6^(@))+1/(sin4 7^(@)sin4 8^(@)) +1/(sin 49^@ sin 50^@)+...+1/(sin 133^@ sin 134^@)` is equal to

Options

A. `sec(1^(@))`
B. `cosec(1^(@))`
C. `cot(1^(@))`
D. None of these

Answer

To solve the given problem, we need to evaluate the sum:

\[ S = \frac{1}{\sin 45^\circ \sin 46^\circ} + \frac{1}{\sin 47^\circ \sin 48^\circ} + \frac{1}{\sin 49^\circ \sin 50^\circ} + \ldots + \frac{1}{\sin 133^\circ \sin 134^\circ} \]

### Step 1: Rewrite the sum in a more manageable form
We can express the sum \(S\) as:

\[ S = \sum_{n=45}^{134} \frac{1}{\sin n^\circ \sin (n+1)^\circ} \]

### Step 2: Use the identity for sine
We can use the identity:

\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \]

Thus, we can rewrite each term in the sum:

\[ \sin n^\circ \sin (n+1)^\circ = \frac{1}{2} [\cos (n - (n+1)) - \cos (n + (n+1))] = \frac{1}{2} [\cos (-1) - \cos (2n + 1)] \]

This gives us:

\[ \frac{1}{\sin n^\circ \sin (n+1)^\circ} = \frac{2}{\cos 1^\circ - \cos (2n + 1)} \]

### Step 3: Substitute back into the sum
Now substituting this back into our sum:

\[ S = \sum_{n=45}^{134} \frac{2}{\cos 1^\circ - \cos (2n + 1)} \]

### Step 4: Simplifying the sum
This sum can be complex, but we can use the telescoping nature of cotangent functions. We can express:

\[ \frac{1}{\sin n^\circ \sin (n+1)^\circ} = \cot n^\circ - \cot (n+1)^\circ \]

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

\[ S = \sum_{n=45}^{134} (\cot n^\circ - \cot (n+1)^\circ) \]

### Step 5: Evaluate the telescoping series
The above sum is telescoping, meaning that most terms will cancel out:

\[ S = \cot 45^\circ - \cot 135^\circ \]

### Step 6: Calculate the cotangent values
We know:

\[ \cot 45^\circ = 1 \quad \text{and} \quad \cot 135^\circ = -1 \]

Thus, we have:

\[ S = 1 - (-1) = 1 + 1 = 2 \]

### Step 7: Final expression
Now, we multiply by \( \frac{1}{\sin 1^\circ} \):

\[ S = \frac{2}{\sin 1^\circ} \]

This can be expressed as:

\[ S = 2 \csc 1^\circ \]

### Conclusion
Thus, the final answer is:

\[ \boxed{2 \csc 1^\circ} \]