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Question Number 115798 by aye48 last updated on 28/Sep/20

Commented by bemath last updated on 29/Sep/20

$$\sin \alpha .\cos \beta =\frac{9}{10}$$$$\Rightarrow 2\sin \alpha .\cos \beta =\frac{9}{5}$$$$\Rightarrow \sin (\alpha +\beta )+\sin (\alpha -\beta )=\frac{9}{5}$$$$\Rightarrow \sin (\alpha +\beta )=\frac{9}{5}-\frac{4}{5}=1$$

Answered by TANMAY PANACEA last updated on 29/Sep/20

$$sin\alpha cos\beta -cos\alpha sin\beta =\frac{4}{5}$$$$cos\alpha sin\beta =\frac{9}{10}-\frac{4}{5}=\frac{1}{10}$$$$sin(\alpha -\beta )=\frac{4}{5}=sin{53}^o$$$$sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta =\frac{9}{10}+\frac{1}{10}=sin{90}^o=1$$$$\alpha +\beta =90$$$$\alpha -\beta =53$$$$\alpha =\frac{90+53}{2}=\frac{143}{2}$$$$\beta =\frac{90-53}{2}=\frac{37}{2}$$$$\frac{tan\alpha }{tan\beta }=\frac{sin\alpha cos\beta }{sin\beta cos\alpha }=\frac{9}{10}\times \frac{10}{1}=9$$

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