In a `DeltaABC`, if `B=90^(@)`, then `tan^(2).(A)/(2)`=

Options

A. `(a-b)/(a+b)`
B. `(b-c)/(b+c)`
C. `(c-a)/(c+a)`
D. `(b+c)/(b-c)`

Explanation

To solve the problem, we need to find the value of \( \tan^2\left(\frac{A}{2}\right) \) in triangle \( ABC \) where angle \( B = 90^\circ \).

### Step-by-Step Solution:

1. **Identify the Triangle**:
Since \( B = 90^\circ \), triangle \( ABC \) is a right triangle with \( A \) and \( C \) as the other two angles.

2. **Use the Half-Angle Formula**:
The half-angle formula for tangent is given by:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} \]

3. **Express Sine and Cosine in Terms of Angle A**:
Using the half-angle formulas:
\[ \sin\left(\frac{A}{2}\right) = \sqrt{\frac{1 - \cos A}{2}} \]
\[ \cos\left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{2}} \]

4. **Substituting into the Tangent Formula**:
Substitute the expressions for sine and cosine into the tangent formula:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sqrt{\frac{1 - \cos A}{2}}}{\sqrt{\frac{1 + \cos A}{2}}} \]

5. **Simplifying the Expression**:
This simplifies to:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \]

6. **Squaring the Tangent**:
Now, we square the tangent to find \( \tan^2\left(\frac{A}{2}\right) \):
\[ \tan^2\left(\frac{A}{2}\right) = \frac{1 - \cos A}{1 + \cos A} \]

7. **Finding Cosine of Angle A**:
In triangle \( ABC \), we can express \( \cos A \) in terms of the sides of the triangle. If we denote the sides opposite to angles \( A \), \( B \), and \( C \) as \( a \), \( b \), and \( c \) respectively, then:
\[ \cos A = \frac{b}{c} \]

8. **Substituting \( \cos A \) into the Expression**:
Substitute \( \cos A \) into the squared tangent expression:
\[ \tan^2\left(\frac{A}{2}\right) = \frac{1 - \frac{b}{c}}{1 + \frac{b}{c}} = \frac{c - b}{c + b} \]

9. **Final Result**:
Thus, the final expression for \( \tan^2\left(\frac{A}{2}\right) \) is:
\[ \tan^2\left(\frac{A}{2}\right) = \frac{c - b}{c + b} \]