Let `f (theta)=(1)/(1+(tan theta)^(2021))`, then the value of `sum_(theta=1^(@))^(89^(@))f(theta)` is equal to

Options

A. 45
B. 44
C. `(89)/(2)`
D. `(91)/(2)`

Answer

To solve the problem, we need to evaluate the sum \( S = \sum_{\theta=1}^{89} f(\theta) \), where \( f(\theta) = \frac{1}{1 + (\tan \theta)^{2021}} \).

### Step 1: Understand the function
The function \( f(\theta) \) is defined as:
\[ f(\theta) = \frac{1}{1 + (\tan \theta)^{2021}} \]

### Step 2: Use the identity for complementary angles
Notice that:
\[ f(90^\circ - \theta) = \frac{1}{1 + (\tan(90^\circ - \theta))^{2021}} = \frac{1}{1 + (\cot \theta)^{2021}} \]
Since \( \cot \theta = \frac{1}{\tan \theta} \), we can rewrite \( f(90^\circ - \theta) \):
\[ f(90^\circ - \theta) = \frac{1}{1 + \left(\frac{1}{\tan \theta}\right)^{2021}} = \frac{1}{1 + \frac{1}{(\tan \theta)^{2021}}} = \frac{(\tan \theta)^{2021}}{(\tan \theta)^{2021} + 1} \]

### Step 3: Sum \( f(\theta) + f(90^\circ - \theta) \)
Now, let's find \( f(\theta) + f(90^\circ - \theta) \):
\[ f(\theta) + f(90^\circ - \theta) = \frac{1}{1 + (\tan \theta)^{2021}} + \frac{(\tan \theta)^{2021}}{(\tan \theta)^{2021} + 1} \]
Finding a common denominator:
\[ = \frac{1 + (\tan \theta)^{2021}}{1 + (\tan \theta)^{2021}} = 1 \]

### Step 4: Pair the terms in the sum
From \( \theta = 1^\circ \) to \( 89^\circ \), we can pair the terms:
- \( f(1^\circ) + f(89^\circ) = 1 \)
- \( f(2^\circ) + f(88^\circ) = 1 \)
- ...
- \( f(44^\circ) + f(46^\circ) = 1 \)

The middle term, \( f(45^\circ) \), is:
\[ f(45^\circ) = \frac{1}{1 + (\tan 45^\circ)^{2021}} = \frac{1}{1 + 1^{2021}} = \frac{1}{2} \]

### Step 5: Count the pairs and calculate the sum
There are \( 44 \) pairs (from \( 1^\circ \) to \( 44^\circ \)), each summing to \( 1 \), plus the middle term:
\[ S = 44 \cdot 1 + \frac{1}{2} = 44 + \frac{1}{2} = 44.5 \]

### Final Answer
Thus, the value of \( S \) is:
\[ \boxed{44.5} \]