Question Number 146044 by mathdanisur last updated on 10/Jul/21 | ||
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$${Simplify}: \\ $$$$\frac{{sin}^{\mathrm{3}} \alpha}{\mathrm{1}-{cos}\alpha}\:+\:\frac{{cos}^{\mathrm{3}} \alpha}{{sin}\alpha+\mathrm{1}}\:=\:? \\ $$ | ||
Answered by liberty last updated on 10/Jul/21 | ||
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$$\:\Rightarrow\:\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\mathrm{sin}\:\alpha}{\mathrm{1}−\mathrm{cos}\:\alpha}\:+\:\frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha\right)\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{sin}\:\alpha}\: \\ $$$$=\:\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)+\mathrm{cos}\:\alpha\left(\mathrm{1}−\mathrm{sin}\:\alpha\right) \\ $$$$=\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha \\ $$ | ||
Commented by mathdanisur last updated on 10/Jul/21 | ||
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$${thankyou}\:{Ser} \\ $$ | ||