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Question Number 147809 by mathdanisur last updated on 23/Jul/21
((sin^6 (𝛂) + cos^6 (𝛂) - 1)/(sin^4 (𝛂) - sin^2 (𝛂))) = ?
$$\frac{\boldsymbol{{sin}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:+\:\boldsymbol{{cos}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:-\:\mathrm{1}}{\boldsymbol{{sin}}^{\mathrm{4}} \left(\boldsymbol{\alpha}\right)\:-\:\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{\alpha}\right)}\:=\:? \\ $$
Answered by gsk2684 last updated on 23/Jul/21
(sin^2 α)^3 +(cos^2 α)^3   =(sin^2 α+cos^2 α)(sin^4 α−sin^2 α cos^2 α+cos^4 α)  =(1)((sin^2 α+cos^2 α)^2 −3sin^2 α cos^2 α)  =1−3sin^2 α cos^2 α  answer  ((−3sin^2 α cos^2 α)/(sin^2 α (sin^2 α−1)))=3
$$\left(\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{sin}\:^{\mathrm{4}} \alpha−\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{4}} \alpha\right) \\ $$$$=\left(\mathrm{1}\right)\left(\left(\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{2}} \alpha\right)^{\mathrm{2}} −\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha\right) \\ $$$$=\mathrm{1}−\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha \\ $$$${answer} \\ $$$$\frac{−\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha}{\mathrm{sin}\:^{\mathrm{2}} \alpha\:\left(\mathrm{sin}\:^{\mathrm{2}} \alpha−\mathrm{1}\right)}=\mathrm{3} \\ $$
Commented by mathdanisur last updated on 23/Jul/21
Thank you Sir  Answer:  3 or −3.?
$${Thank}\:{you}\:{Sir} \\ $$$${Answer}:\:\:\mathrm{3}\:{or}\:−\mathrm{3}.? \\ $$
Commented by puissant last updated on 23/Jul/21
3 because ((−3sin^2 α cos^2 α)/(sin^2 α(sin^2 α−1)))  =((−3cos^2 α)/(−(1−sin^2 α)))=((3(1−sin^2 α))/((1−sin^2 α)))=3...
$$\mathrm{3}\:\mathrm{because}\:\frac{−\mathrm{3sin}^{\mathrm{2}} \alpha\:\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{sin}^{\mathrm{2}} \alpha\left(\mathrm{sin}^{\mathrm{2}} \alpha−\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{3cos}^{\mathrm{2}} \alpha}{−\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \alpha\right)}=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \alpha\right)}{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \alpha\right)}=\mathrm{3}… \\ $$
Commented by mathdanisur last updated on 23/Jul/21
thank you Sir
$${thank}\:{you}\:{Sir} \\ $$

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