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Question Number 157925 by cortano last updated on 29/Oct/21

Answered by FongXD last updated on 30/Oct/21

$$\mathrm{Given}:\frac{\sin \alpha \cos \alpha }{\sin \beta \cos \beta }=\frac{8}{5}\left(1\right)\mathrm{and}\frac{\sin \alpha \cos \beta }{\sin \beta \cos \alpha }=4\left(2\right)$$$$\iff \left(1\right)\cdot \left(2\right):\frac{{\sin }^2\alpha }{{\sin }^2\beta }=\frac{32}{5}\mathrm{and}\left(1\right)\mathbf{÷}\left(2\right):\frac{{\cos }^2\alpha }{{\cos }^2\beta }=\frac{2}{5}$$$$\iff 5-{5\sin }^2\alpha =2-{2\sin }^2\beta$$$$\mathrm{but}{5\sin }^2\alpha ={32\sin }^2\beta$$$$\iff 5-{32\sin }^2\beta =2-{2\sin }^2\beta$$$$\iff {30\sin }^2\beta =3$$$$\iff \sin {\beta }^2=\frac{1}{10},\Rightarrow {\cos }^2\beta =\sqrt{\frac{99}{100}}=\frac{3\sqrt{11}}{10}$$$$\mathrm{we}\mathrm{get}:\tan \alpha +\tan \beta =5\tan \beta =\pm 5\sqrt{\frac{{\sin }^2\beta }{{\cos }^2\beta }}=\pm \frac{5}{\sqrt{3\sqrt{11}}}$$

Commented by cortano last updated on 30/Oct/21

$$\cos {}^2\beta =1-\sin {}^2\beta =1-\frac{1}{10}=\frac{9}{10}$$$$\cos {}^2\beta =\frac{9}{10}?$$

Commented by FongXD last updated on 30/Oct/21

$$\mathrm{ohh}\mathrm{yeah},\mathrm{my}\mathrm{bad},\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{for}\mathrm{correcting}!!$$

Commented by cortano last updated on 30/Oct/21

$$greatsir$$

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