Menu Close

Author: Tinku Tara

Let-x-y-z-gt-0-x-y-z-3-Prove-That-1-x-2-2x-1-z-2-2z-3-1-y-2-y-9-x-y-z-24-1-3-3-17-3-3-

Question Number 197950 by York12 last updated on 05/Oct/23 $${Let}\:{x},{y},{z}>\mathrm{0}\:,\:{x}+{y}+{z}=\mathrm{3}\:{Prove}\:{That}\:: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}+\frac{\mathrm{1}}{\:\sqrt{{z}^{\mathrm{2}} +\mathrm{2}{z}}}+\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{{y}+\mathrm{2}}−\frac{{y}}{\mathrm{9}}\right)+\frac{\sqrt[{\mathrm{3}}]{\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}+\mathrm{24}}}{\:\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{17}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

tan-pi-12-sin-sin-pi-12-cos-cos-pi-12-

Question Number 197944 by liuxinnan last updated on 05/Oct/23 $${tan}\frac{\pi}{\mathrm{12}}=\frac{{sin}\alpha−{sin}\frac{\pi}{\mathrm{12}}}{{cos}\alpha+{cos}\frac{\pi}{\mathrm{12}}} \\ $$$$\alpha=? \\ $$ Answered by MM42 last updated on 05/Oct/23 $${sin}\frac{\pi}{\mathrm{12}}×{cos}\alpha+{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\alpha×{cos}\frac{\pi}{\mathrm{12}}−{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow{sin}\left(\alpha−\frac{\pi}{\mathrm{12}}\right)=\mathrm{2}{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\frac{\pi}{\mathrm{6}} \\…

calculate-L-lim-n-1-1-2-1-1-3-1-1-n-1-n-

Question Number 197947 by mnjuly1970 last updated on 05/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$ Answered by MM42 last updated on…

3-sin-2-tan-cos-2-

Question Number 197937 by mathlove last updated on 05/Oct/23 $$\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} \theta\centerdot{tan}\beta+{cos}^{\mathrm{2}} \beta=? \\ $$ Answered by Frix last updated on 05/Oct/23 $$\mathrm{It}\:\mathrm{is}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is},\:\mathrm{not}\:\mathrm{getting}\:\mathrm{better}\:\mathrm{any}\:\mathrm{way}. \\ $$$${p}=\mathrm{tan}\:\theta\:\wedge{q}=\mathrm{tan}\:\beta\:\rightarrow\:\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} {q}}{{p}^{\mathrm{2}}…

Question-197917

Question Number 197917 by sonukgindia last updated on 04/Oct/23 Commented by Frix last updated on 04/Oct/23 $$\mathrm{I}\:\mathrm{have}\:\mathrm{more}\:\mathrm{joy}\:\mathrm{with}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one},\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{one} \\ $$$$\mathrm{is}\:\mathrm{boring}. \\ $$ Answered by…