Question:
In a \( \Delta ABC \), \( \angle A=\frac{\pi}{2} \). If \( \tan\frac{B}{2} \) and \( \tan\frac{C}{2} \) are the roots of \( 2x^{2}-bx-c=0 \), then what is the value of \( (b-c)^{3} \) ________ ?
In a \( \Delta ABC \), \( \angle A=\frac{\pi}{2} \). If \( \tan\frac{B}{2} \) and \( \tan\frac{C}{2} \) are the roots of \( 2x^{2}-bx-c=0 \), then what is the value of \( (b-c)^{3} \) ________ ?
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Since A = 90 degrees, B/2 + C/2 = 45 degrees, so the tangent of their sum is 1; use the sum and product of the roots.
Updated On: Nov 12, 2026
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Correct Answer: 8
Solution and Explanation
As \( A=\frac{\pi}{2} \), \( \frac B2+\frac C2=\frac{\pi}{4} \), hence \( \tan\left(\frac B2+\frac C2\right)=1 \).
For the roots of \( 2x^2-bx-c=0 \): sum = \( \frac b2 \), product = \( -\frac c2 \).
\( \tan\left(\frac B2+\frac C2\right)=\frac{\text{sum}}{1-\text{product}}=\frac{b/2}{1+c/2}=\frac{b}{2+c}=1 \).
So \( b=c+2 \), i.e. \( b-c=2 \) and \( (b-c)^3=8 \).
Answer: 8
For the roots of \( 2x^2-bx-c=0 \): sum = \( \frac b2 \), product = \( -\frac c2 \).
\( \tan\left(\frac B2+\frac C2\right)=\frac{\text{sum}}{1-\text{product}}=\frac{b/2}{1+c/2}=\frac{b}{2+c}=1 \).
So \( b=c+2 \), i.e. \( b-c=2 \) and \( (b-c)^3=8 \).
Answer: 8
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