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Question Number 86948 by gny last updated on 01/Apr/20

$$\frac{-sin\alpha +3cos\alpha }{4cos\alpha +3sin\alpha }$$$$$$$$tg\alpha =-3$$$$plshelpmesir$$

Commented by Prithwish Sen 1 last updated on 01/Apr/20

$$\mathrm{divide}{\mathrm{n}}_{\mathrm{r}}\mathrm{and}{\mathrm{d}}_{\mathrm{r}}\mathrm{by}\cos \alpha \mathrm{we}\mathrm{get}$$$$\frac{-\tan \alpha +3}{4+3\tan \alpha }=\frac{3+3}{4-9}=-\frac{6}{5}$$

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