Multiple choice

Given that Tan A, Tan B are the roots of the equation ${x^2} - bx -c = 0$ then the value of ${\sin ^2}\left( {A + B} \right)$ is

  1. $\dfrac{b}{{{{\left( {b + c} \right)}^2}}}$
  2. $\dfrac{{{b^2}}}{{{b^2} + {c^2}}}$
  3. $\dfrac{{{b^2}}}{{{c^2} + {{\left( {1 - b} \right)}^2}}}$
  4. $\dfrac{{{b^2}}}{{{b^2} + {{\left( {1 - c} \right)}^2}}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

tan A + tan B = b, tan A * tan B = -c. tan(A+B) = (tan A + tan B) / (1 - tan A tan B) = b / (1 + c). sin^2(A+B) = tan^2(A+B) / (1 + tan^2(A+B)) = [b^2 / (1+c)^2] / [1 + b^2 / (1+c)^2] = b^2 / ((1+c)^2 + b^2). The provided option D uses (1-c)^2, which implies a sign difference in the original equation or identity.

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