Question

If $\tan \alpha$ and $\tan \beta$ are the roots of $x^2-px+q=0$, then find the value of ${\sin }^2(\alpha +\beta )$.

Answer 1

Given, $x^2-px+q=0$

Let $\tan \alpha$ and $\tan \beta$ be the roots of the equation

Sum of the roots$=\tan \alpha +\tan \beta =p$

Product of the roots$=\tan \alpha \tan \beta =q$

We know that $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{p}{1-q}$

From the right angled triangle,

$\tan (\alpha +\beta )=\frac{p}{1-q}$

$\Rightarrow oppositeside=p$ and $adjacentside=1-q$

Hence hypotenuse$=\sqrt{p^2+{(1-q)}^2}=\sqrt{p^2+q^2+1-2q}$

Now,$\sin (\alpha +\beta )=\frac{oppositeside}{hypotenuse}=\frac{p}{\sqrt{p^2+q^2+1-2q}}$

$\therefore {\sin }^2(\alpha +\beta )=\frac{p^2}{p^2+q^2+1-2q}$