If $$ an \alpha$$ and $$ an \beta$$ are the two real roots of the function $$x^2+px+q=0$$, find the value of $$\dfrac{\sin (\alpha +\beta) }{\cos(\alpha -\beta )} $$

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If $$	an \alpha$$ and $$	an \beta$$ are the two real roots of the function $$x^2+px+q=0$$, find the value of $$\dfrac{\sin (\alpha +\beta) }{\cos(\alpha -\beta )}    $$

Uniteasy Admin · Accepted Answer

Since $$	an \alpha$$ and $$	an \beta$$ are the two real roots of the function $$x^2+px+q=0$$
$$	an \alpha 	an \beta =q $$
$$	an \alpha +	an \beta =p $$
$$\dfrac{\sin (\alpha +\beta) }{\cos(\alpha -\beta )}  $$ 
$$=\dfrac{\sin \alpha\cos \beta +\cos \alpha \sin \beta}{\cos \alpha\cos \beta +\sin \alpha \sin \beta} $$
$$=\dfrac{	an \alpha +	an \beta  }{1+	an \alpha 	an \beta  } $$
$$=\dfrac{p}{1+q} $$

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