Question Number 19385 by NEC last updated on 10/Aug/17 $${tan}^{\mathrm{2}} \beta=−\mathrm{1} \\ $$$$ \\ $$$${find}\:\beta….\:{lets}\:{solve}\:{for}\:{fun} \\ $$ Commented by NEC last updated on 10/Aug/17 $${please}\:{help}…
Question Number 19367 by Tinkutara last updated on 10/Aug/17 $$\mathrm{If}\:\alpha,\:\beta,\:\gamma\:\mathrm{and}\:\delta\:\mathrm{are}\:\mathrm{four}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:\left(\theta\:+\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:=\:\mathrm{3}\:\mathrm{tan}\:\mathrm{3}\theta,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\Sigma\mathrm{tan}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:=\:−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\delta\:=\:−\mathrm{3} \\ $$ Answered by ajfour…
Question Number 84868 by jagoll last updated on 17/Mar/20 $$\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{x}\right)}\:=\:−\mathrm{cos}\:\mathrm{x}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{x}−\pi\right) \\ $$ Answered by john santu last updated on 17/Mar/20 $$\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{x}\right)\:=\:−\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{sin}\:\left(\mathrm{x}−\pi\right)\:=\:−\mathrm{sin}\:\mathrm{x} \\…
Question Number 19292 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:…\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$ Commented by prakash jain last updated on 09/Aug/17 $$\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following} \\…
Question Number 19291 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:. \\ $$ Answered by Tinkutara last updated on 16/Aug/17 $$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{sin}\:\mathrm{48}°\:+\:\mathrm{sin}\:\mathrm{24}°}{\mathrm{sin}\:\mathrm{48}°\:\mathrm{sin}\:\mathrm{24}°} \\…
Question Number 19243 by Tinkutara last updated on 07/Aug/17 $$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{opposite}\:\mathrm{the} \\ $$$$\mathrm{angles}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{respectively}\:\mathrm{of}\:\mathrm{a} \\ $$$$\Delta\mathrm{ABC}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{a}\:+\:{b}\:=\:{kc} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:=\:{k}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}. \\ $$ Commented by Tinkutara last updated…
Question Number 19122 by Tinkutara last updated on 05/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:{r}_{\mathrm{1}} \:+\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:=\:\mathrm{4}{R}\:+\:{r} \\ $$ Answered by ajfour last updated on 05/Aug/17 $$\mathrm{2Rsin}\:\mathrm{A}=\mathrm{a} \\ $$$$\mathrm{r}_{\mathrm{1}}…
Question Number 84651 by ajfour last updated on 14/Mar/20 $${A}\:{person}\:{stands}\:{in}\:{the}\:{diagonal} \\ $$$${produced}\:{of}\:{the}\:{square}\:{base}\:{of}\:{a} \\ $$$${church}\:{tower},\:{at}\:{a}\:{distance}\:\mathrm{2}{a} \\ $$$${from}\:{it},\:{and}\:{observes}\:{the}\:{angle} \\ $$$${of}\:{elevation}\:{of}\:{each}\:{of}\:{the}\:{two} \\ $$$${outer}\:{corners}\:{of}\:{the}\:{top}\:{of}\:{the} \\ $$$${tower}\:{to}\:{be}\:\mathrm{30}°,\:{while}\:{that}\:{of}\:{the} \\ $$$${nearest}\:{corner}\:{is}\:\mathrm{45}°.\:{Find}\:{the} \\…
Question Number 19097 by Tinkutara last updated on 04/Aug/17 $$\mathrm{If}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}\:+\:{x}\right)\:=\:\mathrm{tan}^{\mathrm{3}} \:\left(\frac{\pi}{\mathrm{4}}\:+\:\alpha\right)\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cosec}\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\alpha\:+\:\mathrm{sin}^{\mathrm{3}} \:\mathrm{2}\alpha} \\ $$ Answered by 951172235v last updated on 01/Feb/19 $$\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}\:\:=\:\:\left(\frac{\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha}\right)^{\mathrm{3}}…
Question Number 19055 by Tinkutara last updated on 03/Aug/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{three}\:\mathrm{escribed}\:\mathrm{circles} \\ $$$$\mathrm{in}\:\mathrm{term}\:\mathrm{of}\:\mathrm{inradius},\:\mathrm{circumradius}\:\mathrm{and} \\ $$$$\mathrm{semiperimeter}. \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on 05/Aug/17…