Question

If \(\tan A\) and \(\tan B\) are the roots of the equation \(x^2 - ax + b = 0\), then the value of \(\sin^2(A + B)\) is

• A. \(\frac{a^2}{a^2 + (1-b)^2}\)
• B. \(\frac{a^2}{a^2 + b^2}\)
• C. \(\frac{a^2}{(a+b)^2}\)
• D. \(\frac{a^2}{b^2 + (1-a)^2}\)

Hint:

\(\sin^2\theta = \frac{\tan^2\theta}{1+\tan^2\theta}\).

Answer 1

The Correct Option is A

To find the value of \(\sin^2(A + B)\) given that \(\tan A\) and \(\tan B\) are the roots of the quadratic equation \(x^2 - ax + b = 0\), we will use the properties of roots and trigonometric identities.

1.  Using Vieta's formulas, we know:
   • The sum of roots: \(\tan A + \tan B = a\)
   • The product of roots: \(\tan A \cdot \tan B = b\)
2. We want to find \(\sin^2(A + B)\). Using the trigonometric identity for sine of a sum: 

\[\sin(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{a}{1-b}\]

1. Then, \(\sin^2(A+B)\) becomes: 

\[\sin^2(A + B) = \left(\frac{a}{1-b}\right)^2 = \frac{a^2}{(1-b)^2}\]

1. Evaluating this within the options, the correct expression matches the first option: 

\[\frac{a^2}{a^2 + (1-b)^2}\]

Thus, the correct answer is \(\frac{a^2}{a^2 + (1-b)^2}\).