Question

If tan A and tan B are the roots of the quadratic equation $x^2$ -ax+b=0, then the value of $si{n}^2$ (A+B) is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We want to find $si{n}^2$ (A+B). If we know tan (A+B) or cos

(A+B) or sin (A+B) or any other basic trigonometric ratio of the angle A+B, we can find $si{n}^2$ (A+B) easily. Since we are given tanA and tanB are the roots of a quadratic education, we can find tanA+tanB and tanAtanB. Once we have these two, we can find tan (A+B).

TanA+tanB = a

TanAtanB = b

$\Rightarrow$ tan (A+B) = $\frac{tanA+tanB}{1-tanAtanB}$

= $\frac{a}{1-b}$

We will construct a $△$ and proceed

Sin (A+B) = $\frac{a}{\sqrt{a^2+(1-b{)}^2}}$
$\Rightarrow si{n}^2(A+B)=\frac{a^2}{a^2+(1-b{)}^2}$

Key steps/concepts : (1) Finding tan (A+B)