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Question Number 128068 by benjo_mathlover last updated on 04/Jan/21

$$\mathrm{Given}{25\mathrm{x}}^2-30\mathrm{x}+7=0\mathrm{has}\mathrm{the}$$ $$\mathrm{roots}\mathrm{are}\cos \alpha \mathrm{and}\cos \beta .\mathrm{If}\cos \alpha -\cos \beta >0$$ $$\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\tan \left(\frac{\alpha +\beta }{2}\right).\tan \left(\frac{\alpha -\beta }{2}\right)=?$$

Answered by bemath last updated on 04/Jan/21

$$\mathrm{let}\frac{\alpha +\beta }{2}=\mathrm{x}\wedge \frac{\alpha -\beta }{2}=\mathrm{y}$$ $$\left(\mathrm{i}\right)\tan \mathrm{x}.\tan \mathrm{y}=\frac{2\sin \mathrm{xsin}\mathrm{y}}{2\cos \mathrm{xcos}\mathrm{y}}$$ $$=\frac{\cos (\mathrm{x}-\mathrm{y})-\cos (\mathrm{x}+\mathrm{y})}{\cos (\mathrm{x}+\mathrm{y})+\cos (\mathrm{x}-\mathrm{y})}$$ $$=\frac{\cos \beta -\cos \alpha }{\cos \alpha +\cos \beta }=\frac{\cos \beta -\cos \alpha }{\left(\frac{6}{5}\right)}$$ $$=-\frac{5}{6}\sqrt{{(\cos \alpha -\cos \beta )}^2}$$ $$=-\frac{5}{6}\sqrt{{(\cos \alpha +\cos \beta )}^2-4\cos \alpha \cos \beta }$$ $$=-\frac{5}{6}\sqrt{\frac{36}{25}-\frac{28}{25}}=-\frac{5}{6}\times \frac{2\sqrt{2}}{5}$$ $$=-\frac{\sqrt{2}}{3}$$

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$cos\alpha +cos\beta =\frac{30}{25}\Rightarrow cos\alpha -cos\beta =\pm \sqrt{\frac{6^2}{5^2}-\frac{28}{25}}=\sqrt{\frac{8}{25}}$$ $$cos\alpha cos\beta =\frac{7}{25}$$ $$tan\left(\frac{\alpha +\beta }{2}\right)tan\left(\frac{\alpha -\beta }{2}\right)=\frac{cos\beta -cos\alpha }{cos\beta +cos\alpha }=-\frac{5}{6}.\frac{2\sqrt{2}}{5}=-\frac{\sqrt{2}}{3}$$ $$$$

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