Give a right angle triangle ABC with right angled at C. `angleC=90^(@)` and a,b,c are lengths of corresponding sides. `(b gt a)`. `tan(A/2)` will be equal to

Options

A. `(2ab)/(a^(2)-b^(2))`
B. `(2ab)/(b^(2)-a^(2))`
C. `((c-b))/a`
D. `(b^(2)-c^(2))/(2bc)`

Answer

To find the value of \( \tan\left(\frac{A}{2}\right) \) in a right-angled triangle \( ABC \) with the right angle at \( C \), we can follow these steps:

### Step 1: Identify the sides of the triangle
In triangle \( ABC \):
- Let \( a \) be the length of side opposite angle \( A \).
- Let \( b \) be the length of side opposite angle \( B \).
- Let \( c \) be the length of the hypotenuse.

Given that \( b > a \), we can denote:
- \( AC = b \) (the side opposite angle \( A \))
- \( BC = a \) (the side opposite angle \( B \))
- \( AB = c \) (the hypotenuse)

### Step 2: Use the Pythagorean theorem
According to the Pythagorean theorem:
\[ c^2 = a^2 + b^2 \]

### Step 3: Find \( \tan A \)
The tangent of angle \( A \) can be expressed as:
\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} \]

### Step 4: Use the half-angle formula for tangent
The half-angle formula for tangent states:
\[ \tan\left(\frac{A}{2}\right) = \frac{1 - \cos A}{\sin A} \]
We can also express it in terms of \( \tan A \):
\[ \tan\left(\frac{A}{2}\right) = \frac{\tan A}{1 + \sec A} \]

### Step 5: Find \( \sin A \) and \( \cos A \)
Using the definitions of sine and cosine:
\[ \sin A = \frac{a}{c}, \quad \cos A = \frac{b}{c} \]

### Step 6: Substitute into the half-angle formula
Now substituting \( \sin A \) and \( \cos A \) into the half-angle formula:
\[ \tan\left(\frac{A}{2}\right) = \frac{\frac{a}{c}}{1 + \frac{b}{c}} = \frac{a}{c + b} \]

### Step 7: Final expression for \( \tan\left(\frac{A}{2}\right) \)
Thus, we can conclude:
\[ \tan\left(\frac{A}{2}\right) = \frac{a}{b + c} \]

### Conclusion
The value of \( \tan\left(\frac{A}{2}\right) \) is \( \frac{a}{b + c} \).

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