Question

If \tan \alpha and \tan \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 )}

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

Since \tan \alpha and \tan \beta are the two real roots of the function x^2+px+q=0

\tan \alpha \tan \beta =q

\tan \alpha +\tan \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{\tan \alpha +\tan \beta }{1+\tan \alpha \tan \beta }

=\dfrac{p}{1+q}

Steven Zheng posted 1 month ago

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