Question Number 108463 by Sunnatilla last updated on 17/Aug/20 $${arcsin}\left({sin}\mathrm{10}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173983 by mnjuly1970 last updated on 23/Jul/22 $$ \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:\:{prove}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\frac{\:{cos}\left({A}\:\right)}{{a}^{\:\mathrm{3}} }\:+\frac{{cos}\left({B}\right)}{{b}^{\:\mathrm{3}} }\:+\frac{{cos}\left({C}\right)}{{c}^{\:\mathrm{3}} }\:\geqslant\frac{\mathrm{81}}{\mathrm{16}{p}^{\:\mathrm{3}} } \\ $$$$\:\:\:{where}\::\:\:{p}=\:\left({a}+{b}\:+{c}\:\right)/\mathrm{2} \\ $$$$…
Question Number 173893 by azadsir last updated on 20/Jul/22 $$\mathrm{if}\:\int\left(\mathrm{x}\right)\:=\:\mathrm{sinx}\:\mathrm{than}\:\mathrm{prove}\:\mathrm{that}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left\{\int\left(\mathrm{x}\right)^{\mathrm{4}} \right\}\:+\:\left\{\int\left(\mathrm{x}\right)\right\}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$ Commented by MJS_new last updated on 20/Jul/22 $$\int\:\mathrm{is}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{sign}… \\…
Question Number 173894 by azadsir last updated on 20/Jul/22 $$\mathrm{If}\:\:\mathrm{secA}\:−\:\mathrm{tanA}\:=\:\mathrm{Q}\:\mathrm{than}\:\mathrm{prove}\:\mathrm{that},\: \\ $$$$\:\:\:\:\:\:\:\mathrm{cosecA}\:=\:\frac{\mathrm{1}\:+\:\mathrm{Q}^{\mathrm{2}} }{\mathrm{1}\:−\:\mathrm{Q}^{\mathrm{2}} }\: \\ $$ Commented by cortano1 last updated on 20/Jul/22 $${Q}=\frac{\mathrm{1}−\mathrm{sin}\:{A}}{\mathrm{cos}\:{A}}\:\Rightarrow{Q}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}}…
Question Number 173888 by azadsir last updated on 20/Jul/22 $$\mathrm{Prove}\:\mathrm{that},\:\frac{\mathrm{13x}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{13x}}+\mathrm{1}}\:=\:\sqrt{\mathrm{13}}\:\mathrm{if},\:\mathrm{x}\:=\:\sqrt{\mathrm{13}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$ Answered by a.lgnaoui last updated on 20/Jul/22 $$ \\ $$$$\frac{\mathrm{13}{x}}{\left({x}^{\mathrm{2}} −\sqrt{\mathrm{13}}{x}+\mathrm{1}\right)}=\frac{\mathrm{13}{x}}{\:\sqrt{\mathrm{13}}{x}}\Rightarrow{x}^{\mathrm{2}} −\sqrt{\mathrm{13}}{x}+\mathrm{1}=\sqrt{\mathrm{13}}{x}…
Question Number 42673 by Tawa1 last updated on 31/Aug/18 Answered by OrestesDante last updated on 15/Sep/18 $$\frac{{p}}{{q}}=\frac{{sin}\mathrm{2}\alpha+{sin}\mathrm{2}\beta}{{cos}\mathrm{2}\alpha+{cos}\mathrm{2}\beta} \\ $$$$\frac{\mathrm{2}{sin}\left(\frac{\mathrm{2}\alpha+\mathrm{2}\beta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\alpha−\mathrm{2}\beta}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\frac{\mathrm{2}\alpha+\mathrm{2}\beta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\alpha−\mathrm{2}\beta}{\mathrm{2}}\right)}=\frac{{p}}{{q}} \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{p}}{{q}} \\ $$$$>>>>>>>>>>>>>>>> \\ $$$$…
Question Number 108145 by bemath last updated on 15/Aug/20 $$\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\Pi} \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\:>\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\: \\ $$$$\:\left(\mathrm{2}\right)\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left(\mathrm{3}\right)\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$ Commented by…
Question Number 108085 by bemath last updated on 14/Aug/20 $$\:\:\:\:\:\frac{\Cup\mathcal{B}{e}\mathcal{M}{ath}\Cup}{\infty} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{75}°−\mathrm{cos}\:^{\mathrm{8}} \mathrm{75}°\:= \\ $$ Commented by bemath last updated on 14/Aug/20 $${thank}\:{you}\:{both} \\…
Question Number 42520 by gyugfeet last updated on 27/Aug/18 $${cos}^{\mathrm{3}} \:{A}.{sin}\mathrm{3}{A}+{sinA}.{cos}\mathrm{3}{A}=\frac{\mathrm{3}}{\mathrm{4}}{sin}\mathrm{4}{A} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18 $${cos}\mathrm{3}{A}=\frac{\mathrm{4}{cos}^{\mathrm{3}} {A}−\mathrm{3}{cosA}}{} \\ $$$$\left(\frac{\mathrm{3}{cosA}+{cos}\mathrm{3}{A}}{\mathrm{4}}\right){sin}\mathrm{3}{A}+{sinAcos}\mathrm{3}{A} \\…
Question Number 42519 by gyugfeet last updated on 27/Aug/18 $${cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}={cotA}−{cot}\mathrm{8}{A}\left({prlve}\:{ghis}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18 $${cotA}−\left({cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$${cotA}−{cose}\mathrm{2}{A}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$$\frac{{cosA}}{{sinA}}−\frac{\mathrm{1}}{\mathrm{2}{sinAcosA}}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\…