Question Number 113781 by Ar Brandon last updated on 15/Sep/20 Answered by Dwaipayan Shikari last updated on 15/Sep/20 $${sin}\frac{\pi}{\mathrm{12}}{sin}\frac{\mathrm{7}\pi}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\frac{\pi}{\mathrm{2}}−{cos}\frac{\mathrm{8}\pi}{\mathrm{12}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Answered by Dwaipayan Shikari…
Question Number 113769 by faysal last updated on 15/Sep/20 $${prove}\:{that},\:\mathrm{tan}\:\left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)°=\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\mathrm{2} \\ $$ Answered by Dwaipayan Shikari last updated on 15/Sep/20 $${tan}\left(\frac{\pi}{\mathrm{24}}\right)=\frac{{sin}\frac{\pi}{\mathrm{24}}}{{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{24}}}{\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{1}−{cos}\frac{\pi}{\mathrm{12}}}{{sin}\frac{\pi}{\mathrm{12}}} \\ $$$${cos}\frac{\pi}{\mathrm{12}}={cos}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\…
Question Number 48066 by rahul 19 last updated on 18/Nov/18 $$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}.\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}.\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}……\infty=? \\ $$ Commented by MJS last updated on 18/Nov/18 $$=\frac{\mathrm{2}}{\pi} \\ $$$$\mathrm{approximated}\:\mathrm{it}\:\mathrm{but}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{prove}\:\mathrm{it} \\ $$…
Question Number 47932 by ajfour last updated on 17/Nov/18 Commented by ajfour last updated on 17/Nov/18 $${Q}.\mathrm{47824} \\ $$ Answered by ajfour last updated on…
Question Number 47836 by Aknabob1 last updated on 15/Nov/18 $${prove}\:{cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta=\mathrm{1} \\ $$ Answered by $@ty@m last updated on 15/Nov/18 $${LHS}={cosec}^{\mathrm{2}} \theta−{cot}\theta{cosec}\theta \\ $$$$={cosec}^{\mathrm{2}} \theta−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}×\frac{\mathrm{1}}{\mathrm{sin}\:\theta}…
Question Number 47831 by somil last updated on 15/Nov/18 Commented by somil last updated on 16/Nov/18 $${i}\:{know}\: \\ $$ Answered by $@ty@m last updated on…
Question Number 178780 by Laftex234 last updated on 21/Oct/22 Answered by MJS_new last updated on 21/Oct/22 $$\mathrm{lhs}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{is}\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{16}}\:\mathrm{at}\:{x}=\pm\mathrm{1}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ Answered by Spillover…
Question Number 178779 by Laftex234 last updated on 21/Oct/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178743 by mathlove last updated on 21/Oct/22 $${f}\left({x}\right)={arctan}\left(\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}}\right)\:\: \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Commented by cortano1 last updated on 21/Oct/22 $$\:\mathrm{tan}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}} \\ $$$$\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 113200 by bobhans last updated on 11/Sep/20 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)=\frac{\pi}{\mathrm{4}} \\ $$ Answered by john santu last updated on 11/Sep/20 $$\left({Q}\right)\:{prove}\:{that}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:\frac{\pi}{\mathrm{4}}.…