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Author: Tinku Tara

If-determinant-x-n-x-n-2-x-n-3-y-n-y-n-2-y-n-3-z-n-z-n-2-z-n-3-x-y-y-z-z-x-1-x-1-y-1-z-then-n-equals-

Question Number 67836 by gunawan last updated on 01/Sep/19 $$\mathrm{If}\:\begin{vmatrix}{{x}^{{n}} }&{{x}^{{n}+\mathrm{2}} }&{{x}^{{n}+\mathrm{3}} }\\{{y}^{{n}} }&{{y}^{{n}+\mathrm{2}} }&{{y}^{{n}+\mathrm{3}} }\\{{z}^{{n}} }&{{z}^{{n}+\mathrm{2}} }&{{z}^{{n}+\mathrm{3}} }\end{vmatrix} \\ $$$$=\:\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left(\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\right), \\ $$$$\mathrm{then}\:{n}\:\mathrm{equals} \\ $$…

Question-193399

Question Number 193399 by Mingma last updated on 12/Jun/23 Answered by MathematicalUser2357 last updated on 10/Sep/23 $${A}=\mathrm{cosec}\:\mathrm{39}°\approx\mathrm{1}.\mathrm{589016} \\ $$$${B}=\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{618034} \\ $$$$\mathrm{So},\:{A}\rangle{B} \\ $$ Terms of…

Question-193391

Question Number 193391 by DAVONG last updated on 12/Jun/23 Answered by aba last updated on 12/Jun/23 $$\left(\mathrm{1}\right)\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{6}\right)−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{ln}\left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{a}}\right)\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{a}}}\:\checkmark \\ $$ Commented…

Question-193385

Question Number 193385 by AnshKumar last updated on 12/Jun/23 Answered by som(math1967) last updated on 12/Jun/23 $${L}.{H}.{S} \\ $$$$=\frac{{sec}\mathrm{8}{A}−\mathrm{1}}{{sec}\mathrm{4}{A}−\mathrm{1}} \\ $$$$=\frac{\frac{\mathrm{1}}{{cos}\mathrm{8}{A}}−\mathrm{1}}{\frac{\mathrm{1}}{{cos}\mathrm{4}{A}}−\mathrm{1}} \\ $$$$=\frac{{cos}\mathrm{4}{A}\left(\mathrm{1}−{cos}\mathrm{8}{A}\right)}{{cos}\mathrm{8}{A}\left(\mathrm{1}−{cos}\mathrm{4}{A}\right)} \\ $$$$=\frac{{cos}\mathrm{4}{A}×\mathrm{2}{sin}^{\mathrm{2}}…

Question-193387

Question Number 193387 by Mingma last updated on 12/Jun/23 Commented by Frix last updated on 12/Jun/23 $$\mathrm{The}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}: \\ $$$${a}=\mathrm{6} \\ $$$${b}=\mathrm{6}\sqrt{\mathrm{7}} \\ $$$${c}=\mathrm{18} \\ $$$$\Rightarrow…

Question-193381

Question Number 193381 by Mingma last updated on 12/Jun/23 Answered by som(math1967) last updated on 12/Jun/23 $$\:\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}\theta\:+\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{2}{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\mathrm{1}−{cos}\mathrm{8}\theta+\mathrm{1}−{cos}\mathrm{4}\theta−\mathrm{1}+{cos}\mathrm{2}\theta=\mathrm{1} \\ $$$${cos}\mathrm{2}\theta−\left({cos}\mathrm{4}\theta+{cos}\mathrm{8}\theta\right)=\mathrm{0} \\…