Question Number 9486 by Joel575 last updated on 10/Dec/16 $$\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{{n}}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{So},\:\frac{\mathrm{1}}{{n}}\:=\:? \\ $$ Answered by mrW last updated on 10/Dec/16 $$\mathrm{tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{{n}}\right)\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}\:}\right) \\ $$$$\frac{\mathrm{tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{n}}\right)}{\mathrm{1}−\mathrm{tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{4}}\right)×\mathrm{tan}\:\left(\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\mathrm{arctan}\:\frac{\mathrm{1}}{{n}}\right)\:}=\mathrm{1} \\…
Question Number 140533 by liberty last updated on 09/May/21 $$\mathrm{A}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{divided} \\ $$$$\mathrm{the}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{into}\:\mathrm{three}\:\mathrm{area}\:\mathrm{of}\:\mathrm{length}\:\mathrm{6},\mathrm{8},\mathrm{10} \\ $$$$\mathrm{units}\:\mathrm{then}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}… \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{64}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\pi^{\mathrm{2}} }\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{36}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\pi^{\mathrm{2}} } \\…
Question Number 74966 by ajfour last updated on 05/Dec/19 $${If}\:\:\mathrm{sin}\:\mathrm{2}{A}=\lambda\mathrm{sin}\:\mathrm{2}{B} \\ $$$${Prove}\:{that}\:\:\frac{\mathrm{tan}\:\left({A}+{B}\right)}{\mathrm{tan}\:\left({A}−{B}\right)}=\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}}\:. \\ $$ Commented by Prithwish sen last updated on 05/Dec/19 $$\frac{\mathrm{sin2A}}{\mathrm{sin2B}}\:=\:\lambda \\ $$$$\frac{\mathrm{sin2A}+\mathrm{sin2B}}{\mathrm{sin2A}−\mathrm{sin2B}}\:=\:\frac{\lambda+\mathrm{1}}{\lambda−\mathrm{1}}…
Question Number 140497 by benjo_mathlover last updated on 08/May/21 $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)=\mathrm{x} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)=? \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{x}\right)=? \\ $$ Commented by mr W last updated…
Question Number 140494 by benjo_mathlover last updated on 08/May/21 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)+\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}\right)=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{x}=? \\ $$ Answered by john_santu last updated on 08/May/21 $$\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{cot}^{−\mathrm{1}}…
Question Number 9414 by Rohit kumar last updated on 06/Dec/16 $${tanA}+{cot}={A}=\:\mathrm{2}{cosec}\mathrm{2}{A} \\ $$ Answered by ridwan balatif last updated on 06/Dec/16 $$\mathrm{tanA}+\mathrm{cotA}=\mathrm{2cosec2A} \\ $$$$\frac{\mathrm{sinA}}{\mathrm{cosA}}+\frac{\mathrm{cosA}}{\mathrm{sinA}}=\mathrm{2cosec2A} \\…
Question Number 140465 by ajfour last updated on 08/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9370 by tawakalitu last updated on 03/Dec/16 Answered by geovane10math last updated on 03/Dec/16 $$\mathrm{There}\:\mathrm{are}\:\mathrm{infinite}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sizes}\:\mathrm{10} \\ $$$$\mathrm{and}\:\mathrm{15}.\:\mathrm{It}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{inclination}\:\mathrm{of}\:{X}\hat {{Z}M}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{w}\:\mathrm{depends}\:\mathrm{of}\: \\ $$$${X}\hat {{Z}M}.…
Question Number 74853 by Raxreedoroid last updated on 01/Dec/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{combine}\:\mathrm{2}{cos}\left(\mathrm{90}°{x}\right)+{cos}\left(\mathrm{180}°{x}\right) \\ $$$$\mathrm{into}\:\mathrm{a}\:\mathrm{form}\:\mathrm{of}\:\:{a}\centerdot{cos}\left({b}\centerdot{x}+{c}\right) \\ $$$$\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}{x}\right)+{cos}\left(\pi{x}\right)\overset{?} {=}{a}\centerdot{cos}\left({bx}+{c}\right) \\ $$ Commented by mr W last updated on 01/Dec/19…
Question Number 140382 by john_santu last updated on 07/May/21 $$\mathrm{sin}\:^{\mathrm{10}} {x}\:+\:\mathrm{cos}\:^{\mathrm{10}} {x}\:=\:\frac{\mathrm{61}}{\mathrm{256}} \\ $$ Answered by MJS_new last updated on 07/May/21 $$\mathrm{sin}\:{x}\:={s}\wedge\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\mathrm{5}{s}^{\mathrm{8}}…