Question

If $\tan \left(A\right)$ and $\tan \left(B\right)$ are the roots of $x^2-ax+b=0$ then the value of ${\sin }^2\left(A+B\right)$ is

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

Answer 1

The correct option is A

$\frac{a^2}{a^2+{\left(1-b\right)}^2}$

Explanation for the correct option

Given that $\tan \left(A\right)$ and $\tan \left(B\right)$ are the roots of $x^2-ax+b=0$

We know that for a quadratic equation $p{x}^2+qx+s=0$,

sum of roots$=\frac{-q}{p}$ and product of roots$=\frac{s}{p}$

$\therefore \tan \left(A\right)+\tan \left(B\right)=a$ and $\tan \left(A\right)\tan \left(B\right)=b$

$\begin{array}{crcll}\Rightarrow & \tan \left(A+B\right)& =& \frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}& \\ & & =& \frac{a}{1-b}& \\ \Rightarrow & {\tan }^2\left(A+B\right)& =& \frac{a^2}{{\left(1-b\right)}^2}& \\ \Rightarrow & {\cot }^2\left(A+B\right)& =& \frac{{\left(1-b\right)}^2}{a^2}& ;\left[\because \cot \left(\theta \right)=\frac{1}{\tan \left(\theta \right)}\right]\\ & & & & \\ \therefore & cose{c}^2\left(A+B\right)-{\cot }^2\left(A+B\right)& =& 1& ;\left[\because cose{c}^2\left(\theta \right)-{\cot }^2\left(\theta \right)=1\right]\\ \Rightarrow & cose{c}^2\left(A+B\right)& =& 1+{\cot }^2\left(A+B\right)& \\ \Rightarrow & & =& 1+\frac{{\left(1-b\right)}^2}{a^2}& ;\left[\because {\cot }^2\left(A+B\right)=\frac{{\left(1-b\right)}^2}{a^2}\right]\\ \Rightarrow & & =& \frac{a^2+{\left(1-b\right)}^2}{a^2}& \\ \Rightarrow & {\sin }^2\left(A+B\right)& =& \frac{a^2}{a^2+{\left(1-b\right)}^2}& ;\left[\because \sin \left(\theta \right)=\frac{1}{\csc \left(\theta \right)}\right]\end{array}$

Hence the correct option is option(A) i.e. $\frac{a^2}{a^2+{\left(1-b\right)}^2}$