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 solve the given problem, we start by analyzing the quadratic equation \(x^2 - ax + b = 0\), where \(\tan A\) and \(\tan B\) are the roots. According to Vieta's formulas, for a quadratic equation \(x^2 - px + q = 0\) with roots \(\alpha\) and \(\beta\):

• \(\alpha + \beta = p\)
• \(\alpha \beta = q\)

Applying Vieta's to our equation, we get:

• \(\tan A + \tan B = a\)
• \(\tan A \cdot \tan B = b\)

Now, using the trigonometric identity for the tangent of the sum of two angles:

\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

Substituting the values from Vieta's:

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

The formula for \(\sin^2\theta\) in terms of \(\tan\theta\) is:

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

Substitute \(\theta = A + B\):

• \(\tan^2(A+B) = \left( \frac{a}{1-b} \right)^2\)
• Therefore, \(\sin^2(A+B) = \frac{\left( \frac{a}{1-b} \right)^2}{1 + \left( \frac{a}{1-b} \right)^2}\)

Simplifying this expression:

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

Multiply the numerator and the denominator by \((1-b)^2\) to clear the fraction in the denominator:

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

Thus, the value of \(\sin^2(A+B)\) is \(\frac{a^2}{a^2 + (1-b)^2}\).

Hence, the correct answer is the first option: \(\frac{a^2}{a^2 + (1-b)^2}\).