Step 1: Understanding the Concept:
We are given a right-angled triangle with \( \angle A = 90^\circ \). The sum of angles is \( 180^\circ \), so \( B + C = 90^\circ \).
We also know the roots of a quadratic equation. We can use the relationships between the roots and coefficients to find the condition.
Step 2: Key Formula or Approach:
If \( r_1, r_2 \) are roots of \( ax^2+bx+c=0 \), then:
Sum of roots: \( r_1 + r_2 = -b/a \)
Product of roots: \( r_1 r_2 = c/a \)
Use trigonometric identity: \( \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \).
Step 3: Detailed Explanation:
In \( \triangle ABC \), since \( A = 90^\circ \), we have:
\[ B + C = 180^\circ - 90^\circ = 90^\circ \]
Divide by 2:
\[ \frac{B}{2} + \frac{C}{2} = 45^\circ \]
Take tangent on both sides:
\[ \tan\left(\frac{B}{2} + \frac{C}{2}\right) = \tan(45^\circ) \]
Using the sum formula for tangent:
\[ \frac{\tan \frac{B}{2} + \tan \frac{C}{2}}{1 - \tan \frac{B}{2} \tan \frac{C}{2}} = 1 \]
We are given that \( \tan \frac{B}{2} \) and \( \tan \frac{C}{2} \) are roots of \( ax^2 + bx + c = 0 \).
So, the sum of the roots is:
\[ \tan \frac{B}{2} + \tan \frac{C}{2} = -\frac{b}{a} \]
And the product of the roots is:
\[ \tan \frac{B}{2} \cdot \tan \frac{C}{2} = \frac{c}{a} \]
Substitute these into the tangent equation:
\[ \frac{-\frac{b}{a}}{1 - \frac{c}{a}} = 1 \]
Simplify the denominator:
\[ \frac{-\frac{b}{a}}{\frac{a - c}{a}} = 1 \]
\[ \frac{-b}{a - c} = 1 \]
Multiply by \( a - c \):
\[ -b = a - c \]
Rearranging terms to match options:
\[ c - b = a \]
\[ c = a + b \]
This is equivalent to \( a + b = c \).
Step 4: Final Answer:
The condition is \( a + b = c \).