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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 Ξ±)^3 +(cos^2 Ξ±)^3 =(sin^2 Ξ±+cos^2 Ξ±)(sin^4 Ξ±βˆ’sin^2 Ξ± cos^2 Ξ±+cos^4 Ξ±) =(1)((sin^2 Ξ±+cos^2 Ξ±)^2 βˆ’3sin^2 Ξ± cos^2 Ξ±) =1βˆ’3sin^2 Ξ± cos^2 Ξ± answer ((βˆ’3sin^2 Ξ± cos^2 Ξ±)/(sin^2 Ξ± (sin^2 Ξ±βˆ’1)))=3

$$\left(\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{sin}\:^{\mathrm{4}} \alphaβˆ’\mathrm{sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{4}} \alpha\right) \\ $$$$=\left(\mathrm{1}\right)\left(\left(\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{cos}\:^{\mathrm{2}} \alpha\right)^{\mathrm{2}} βˆ’\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha\right) \\ $$$$=\mathrm{1}βˆ’\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha \\ $$$${answer} \\ $$$$\frac{βˆ’\mathrm{3sin}\:^{\mathrm{2}} \alpha\:\mathrm{cos}\:^{\mathrm{2}} \alpha}{\mathrm{sin}\:^{\mathrm{2}} \alpha\:\left(\mathrm{sin}\:^{\mathrm{2}} \alphaβˆ’\mathrm{1}\right)}=\mathrm{3} \\ $$

Commented by mathdanisur last updated on 23/Jul/21

Thank you Sir Answer: 3 or βˆ’3.?

$${Thank}\:{you}\:{Sir} \\ $$$${Answer}:\:\:\mathrm{3}\:{or}\:βˆ’\mathrm{3}.? \\ $$

Commented by puissant last updated on 23/Jul/21

3 because ((βˆ’3sin^2 Ξ± cos^2 Ξ±)/(sin^2 Ξ±(sin^2 Ξ±βˆ’1))) =((βˆ’3cos^2 Ξ±)/(βˆ’(1βˆ’sin^2 Ξ±)))=((3(1βˆ’sin^2 Ξ±))/((1βˆ’sin^2 Ξ±)))=3...

$$\mathrm{3}\:\mathrm{because}\:\frac{βˆ’\mathrm{3sin}^{\mathrm{2}} \alpha\:\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{sin}^{\mathrm{2}} \alpha\left(\mathrm{sin}^{\mathrm{2}} \alphaβˆ’\mathrm{1}\right)} \\ $$$$=\frac{βˆ’\mathrm{3cos}^{\mathrm{2}} \alpha}{βˆ’\left(\mathrm{1}βˆ’\mathrm{sin}^{\mathrm{2}} \alpha\right)}=\frac{\mathrm{3}\left(\mathrm{1}βˆ’\mathrm{sin}^{\mathrm{2}} \alpha\right)}{\left(\mathrm{1}βˆ’\mathrm{sin}^{\mathrm{2}} \alpha\right)}=\mathrm{3}... \\ $$

Commented by mathdanisur last updated on 23/Jul/21

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

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