If `tan^(2)B=(1-sinA)/(1+sinA)` then what is the value of A+2B ?

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Explanation

To solve the equation \( \tan^2 B = \frac{1 - \sin A}{1 + \sin A} \) and find the value of \( A + 2B \), we will follow these steps:

### Step 1: Start with the given equation
We have:
\[ \tan^2 B = \frac{1 - \sin A}{1 + \sin A} \]

### Step 2: Multiply both sides by \( 1 + \sin A \)
To eliminate the fraction, we multiply both sides by \( 1 + \sin A \):
\[ \tan^2 B (1 + \sin A) = 1 - \sin A \]

### Step 3: Expand the left side
Expanding the left side gives:
\[ \tan^2 B + \tan^2 B \sin A = 1 - \sin A \]

### Step 4: Rearrange the equation
Now, we will rearrange the equation to group the terms involving \( \sin A \):
\[ \tan^2 B + \tan^2 B \sin A + \sin A = 1 \]
This can be rewritten as:
\[ \tan^2 B + \sin A (1 + \tan^2 B) = 1 \]

### Step 5: Isolate \( \sin A \)
Now, isolate \( \sin A \):
\[ \sin A (1 + \tan^2 B) = 1 - \tan^2 B \]
\[ \sin A = \frac{1 - \tan^2 B}{1 + \tan^2 B} \]

### Step 6: Use the identity for sine
Recall that \( \sin A \) can also be expressed in terms of cosine:
Using the identity \( \tan^2 B = \frac{\sin^2 B}{\cos^2 B} \), we can express \( \tan^2 B \) in terms of sine and cosine:
\[ \sin A = \frac{1 - \frac{\sin^2 B}{\cos^2 B}}{1 + \frac{\sin^2 B}{\cos^2 B}} = \frac{\cos^2 B - \sin^2 B}{\cos^2 B + \sin^2 B} \]
Since \( \cos^2 B + \sin^2 B = 1 \), this simplifies to:
\[ \sin A = \cos 2B \]

### Step 7: Relate \( A \) and \( B \)
From the identity \( \sin A = \cos 2B \), we can use the co-function identity:
\[ A = \frac{\pi}{2} - 2B \]

### Step 8: Find \( A + 2B \)
Now, we can find \( A + 2B \):
\[ A + 2B = \left(\frac{\pi}{2} - 2B\right) + 2B = \frac{\pi}{2} \]

### Final Answer
Thus, the value of \( A + 2B \) is:
\[ \boxed{\frac{\pi}{2}} \]