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Question-56575




Question Number 56575 by Sr@2004 last updated on 18/Mar/19
Commented by Sr@2004 last updated on 18/Mar/19
please solve 12 er 1 no.
$${please}\:{solve}\:\mathrm{12}\:{er}\:\mathrm{1}\:{no}. \\ $$
Commented by Sr@2004 last updated on 20/Mar/19
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Mar/19
u_n =sin^n α+cos^n α  u_4 =sin^4 α+cos^4 α  =(sin^2 +cos^2 α)^2 −2sin^2 αcos^2 α  6u_4 =6−12sin^2 αcos^2 α  p++  u_6 =sin^6 α+cos^6 α  =(sin^2 +cos^2 α)^3 −3sin^2 cos^2 α(sin^2 α+cos^2 α)  =1−3sin^2 αcos^2 α  4u_6 =4−12sin^2 αcos^2 α    6u_4 −4u_6   =6−12sin^2 αco^2 α−4+12sin^ αcos^2 α  so answet is 6−4=2
$${u}_{{n}} ={sin}^{{n}} \alpha+{cos}^{{n}} \alpha \\ $$$${u}_{\mathrm{4}} ={sin}^{\mathrm{4}} \alpha+{cos}^{\mathrm{4}} \alpha \\ $$$$=\left({sin}^{\mathrm{2}} +{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} \alpha{cos}^{\mathrm{2}} \alpha \\ $$$$\mathrm{6}{u}_{\mathrm{4}} =\mathrm{6}−\mathrm{12}{sin}^{\mathrm{2}} \alpha{cos}^{\mathrm{2}} \alpha \\ $$$${p}++ \\ $$$${u}_{\mathrm{6}} ={sin}^{\mathrm{6}} \alpha+{cos}^{\mathrm{6}} \alpha \\ $$$$=\left({sin}^{\mathrm{2}} +{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{3}} −\mathrm{3}{sin}^{\mathrm{2}} {cos}^{\mathrm{2}} \alpha\left({sin}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \alpha\right) \\ $$$$=\mathrm{1}−\mathrm{3}{sin}^{\mathrm{2}} \alpha{cos}^{\mathrm{2}} \alpha \\ $$$$\mathrm{4}{u}_{\mathrm{6}} =\mathrm{4}−\mathrm{12}{sin}^{\mathrm{2}} \alpha{cos}^{\mathrm{2}} \alpha \\ $$$$ \\ $$$$\mathrm{6}{u}_{\mathrm{4}} −\mathrm{4}{u}_{\mathrm{6}} \\ $$$$=\mathrm{6}−\mathrm{12}{sin}^{\mathrm{2}} \alpha{co}^{\mathrm{2}} \alpha−\mathrm{4}+\mathrm{12}{sin}^{} \alpha{cos}^{\mathrm{2}} \alpha \\ $$$${so}\:{answet}\:{is}\:\mathrm{6}−\mathrm{4}=\mathrm{2} \\ $$$$ \\ $$

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