What is the simplified value of `(sqrt(sec^(2)theta+cosec^(2)theta))/(2)` ?

Options

A. `cosec2theta`
B. `sec^(2)theta`
C. `cosecthetasectheta`
D. `tantheta`

Answer

To simplify the expression \(\frac{\sqrt{\sec^2 \theta + \csc^2 \theta}}{2}\), we can follow these steps:

### Step 1: Rewrite secant and cosecant in terms of sine and cosine
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \]
Thus, we can express \(\sec^2 \theta\) and \(\csc^2 \theta\) as:
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} \quad \text{and} \quad \csc^2 \theta = \frac{1}{\sin^2 \theta} \]

### Step 2: Substitute into the expression
Substituting these into the original expression gives us:
\[ \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} \]

### Step 3: Find a common denominator
The common denominator for \(\frac{1}{\cos^2 \theta}\) and \(\frac{1}{\sin^2 \theta}\) is \(\sin^2 \theta \cos^2 \theta\). Thus, we can rewrite the expression as:
\[ \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]

### Step 4: Use the Pythagorean identity
Using the Pythagorean identity, we know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
So, we can simplify:
\[ \sec^2 \theta + \csc^2 \theta = \frac{1}{\sin^2 \theta \cos^2 \theta} \]

### Step 5: Substitute back into the square root
Now we substitute this back into the square root:
\[ \sqrt{\sec^2 \theta + \csc^2 \theta} = \sqrt{\frac{1}{\sin^2 \theta \cos^2 \theta}} = \frac{1}{\sin \theta \cos \theta} \]

### Step 6: Final expression
Now we can substitute this back into the original expression:
\[ \frac{\sqrt{\sec^2 \theta + \csc^2 \theta}}{2} = \frac{1}{2 \sin \theta \cos \theta} \]

### Step 7: Use the double angle identity
Using the double angle identity, we know that:
\[ \sin(2\theta) = 2 \sin \theta \cos \theta \]
Thus, we can rewrite our expression as:
\[ \frac{1}{2 \sin \theta \cos \theta} = \frac{1}{\sin(2\theta)} \]

### Final Answer
The simplified value of \(\frac{\sqrt{\sec^2 \theta + \csc^2 \theta}}{2}\) is:
\[ \frac{1}{\sin(2\theta)} \]