ln a triangle `ABC, B=90^@` then the value of `tan(A/2)=`

Options

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

Answer

To find the value of \( \tan\left(\frac{A}{2}\right) \) in triangle \( ABC \) where \( B = 90^\circ \), we can follow these steps:

### Step 1: Understand the triangle properties
In triangle \( ABC \), since \( B = 90^\circ \), we know that the sum of angles in a triangle is \( 180^\circ \). Therefore, we can express the angles as:
\[ A + B + C = 180^\circ \implies A + 90^\circ + C = 180^\circ \implies A + C = 90^\circ \]
This implies that \( C = 90^\circ - A \).

**Hint:** Remember that the sum of angles in a triangle is always \( 180^\circ \).

### Step 2: Use the half-angle tangent formula
We know the formula for \( \tan\left(\frac{A}{2}\right) \) in terms of the angles of the triangle:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sin A}{1 + \cos A} \]
However, we can also express it using the angles \( B \) and \( C \):
\[ \tan\left(\frac{A}{2}\right) = \frac{B - C}{B + C} \cdot \cot\left(\frac{A}{2}\right) \]

**Hint:** The half-angle formulas can often simplify the problem.

### Step 3: Substitute \( C \)
Since we have \( C = 90^\circ - A \), we can substitute this into our formula:
\[ \tan\left(\frac{A}{2}\right) = \frac{B - (90^\circ - A)}{B + (90^\circ - A)} \]

**Hint:** Substituting known values can simplify the expression.

### Step 4: Simplify the expression
Substituting \( B = 90^\circ \):
\[ \tan\left(\frac{A}{2}\right) = \frac{90^\circ - (90^\circ - A)}{90^\circ + (90^\circ - A)} = \frac{A}{180^\circ - A} \]

**Hint:** Simplifying fractions can lead to clearer results.

### Step 5: Use the identity for tangent
We know that \( \tan\left(\frac{A}{2}\right) \) can also be expressed as:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{B - C}{B + C}} \]
Using \( C = 90^\circ - A \):
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{B - (90^\circ - A)}{B + (90^\circ - A)}} \]

**Hint:** Using square roots can help in finding the final answer.

### Step 6: Final expression
After simplifying, we find:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{B - C}{B + C}} = \sqrt{\frac{90^\circ - (90^\circ - A)}{90^\circ + (90^\circ - A)}} = \sqrt{\frac{A}{180^\circ - A}} \]

Thus, the final result is:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{B - C}{B + C}} \]

### Final Answer
The value of \( \tan\left(\frac{A}{2}\right) \) is \( \sqrt{\frac{B - C}{B + C}} \).