If `f'(x) = tan^(-1)(Sec x + tan x), x in (-pi/2 , pi/2)` and `f(0) = 0` then the value of `f(1)` is

Options

A. `(pi + 1)/4`
B. `(pi - 1)/4`
C. `(pi + 1)/2`
D. 0

Explanation

To solve the problem, we start with the given derivative of the function \( f'(x) = \tan^{-1}(\sec x + \tan x) \) and the initial condition \( f(0) = 0 \). We want to find the value of \( f(1) \).

### Step 1: Simplifying \( f'(x) \)

We know that:
\[ f'(x) = \tan^{-1}(\sec x + \tan x) \]
Using the identity \( \sec x + \tan x = \frac{1 + \sin x}{\cos x} \), we can rewrite:
\[ f'(x) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 2: Recognizing the derivative form

We can also express \( \sec x + \tan x \) in terms of a tangent function:
\[ \sec x + \tan x = \frac{1 + \sin x}{\cos x} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \]
Thus, we have:
\[ f'(x) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) = \frac{\pi}{4} + \frac{x}{2} \]

### Step 3: Integrating \( f'(x) \)

Now we integrate \( f'(x) \):
\[ f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C \]

### Step 4: Using the initial condition

We apply the initial condition \( f(0) = 0 \):
\[ f(0) = \frac{\pi}{4}(0) + \frac{(0)^2}{4} + C = 0 \implies C = 0 \]
Thus, the function simplifies to:
\[ f(x) = \frac{\pi}{4}x + \frac{x^2}{4} \]

### Step 5: Finding \( f(1) \)

Now we calculate \( f(1) \):
\[ f(1) = \frac{\pi}{4}(1) + \frac{(1)^2}{4} = \frac{\pi}{4} + \frac{1}{4} \]
Combining these gives:
\[ f(1) = \frac{\pi + 1}{4} \]

### Final Answer

Thus, the value of \( f(1) \) is:
\[ \frac{\pi + 1}{4} \]

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