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Question-194654

Question Number 194654 by sonukgindia last updated on 12/Jul/23 Answered by MM42 last updated on 12/Jul/23 $$\left(\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{tanx}} \right)^{{tanx}+\mathrm{2}} ={u} \\ $$$$\Rightarrow{u}+\frac{\mathrm{1}}{{u}}=\mathrm{6}\Rightarrow{u}^{\mathrm{2}} −\mathrm{6}{u}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\…

Question-194640

Question Number 194640 by cortano12 last updated on 12/Jul/23 $$\:\:\:\:\underbrace{ } \\ $$ Answered by horsebrand11 last updated on 12/Jul/23 $$\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:=\:\mathrm{y} \\ $$$$\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}}{\mathrm{y}\:\mathrm{sin}\:\mathrm{3y}}\:=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}}…

If-A-a-b-c-b-c-a-c-a-b-and-a-b-c-gt-0-such-that-abc-1-and-A-T-A-I-find-a-3-b-3-c-3-3abc-

Question Number 194642 by horsebrand11 last updated on 12/Jul/23 $$\:\mathrm{If}\:\mathrm{A}=\begin{pmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}\:\:\:\:\:\:\mathrm{c}}\\{\mathrm{b}\:\:\:\:\mathrm{c}\:\:\:\:\:\:\mathrm{a}}\\{\mathrm{c}\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:\mathrm{and}\:\mathrm{a},\mathrm{b},\mathrm{c}\:>\mathrm{0} \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{abc}=\mathrm{1}\:\mathrm{and}\:\mathrm{A}^{\mathrm{T}} .\mathrm{A}=\mathrm{I} \\ $$$$\:\mathrm{find}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}\:. \\ $$ Answered by som(math1967) last updated…

Prove-that-n-IN-k-1-2-n-1-1-sin-2-kpi-2-n-1-2-2n-1-2-3-Give-in-terms-of-n-k-1-2-n-1-1-sin-4-kpi-2-n-1-

Question Number 194638 by Erico last updated on 12/Jul/23 $$\mathrm{Prove}\:\mathrm{that}\:\forall{n}\in\mathrm{IN}^{\ast} \:\:\:\:\: \\ $$$$\:\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)}=\:\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Give}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}\:\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{{sin}^{\mathrm{4}} \left(\frac{{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}}…