Question Number 109282 by mnjuly1970 last updated on 22/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 174779 by mnjuly1970 last updated on 10/Aug/22 Commented by behi834171 last updated on 10/Aug/22 $$\boldsymbol{{sir}}!\:\:\boldsymbol{{your}}\:\boldsymbol{{used}}\:\boldsymbol{{fonts}}\:\boldsymbol{{are}}\:\boldsymbol{{too}}\:\boldsymbol{{small}}. \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{be}}\:\boldsymbol{{friend}}\:\boldsymbol{{with}}\:\boldsymbol{{my}}\:\boldsymbol{{eyes}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{use}}\:\boldsymbol{{larger}}\:\boldsymbol{{fonts}}.\boldsymbol{{thanks}}\:\boldsymbol{{in}}\:\boldsymbol{{advance}}. \\ $$ Commented by…
Question Number 174666 by mnjuly1970 last updated on 07/Aug/22 Answered by behi834171 last updated on 08/Aug/22 $$\boldsymbol{{cos}}\frac{\boldsymbol{{C}}}{\mathrm{2}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{ab}}}} \\ $$$$\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\Psi}=\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\Psi}=\mathrm{1}−\frac{\mathrm{4}\boldsymbol{{ab}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }.\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{ab}}}= \\ $$$$=\mathrm{1}−\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }=…
Question Number 109116 by 675480065 last updated on 21/Aug/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} } \\ $$$${Please}\:{help} \\ $$ Answered by 1549442205PVT last updated on 21/Aug/20 $$…
Question Number 43543 by peter frank last updated on 11/Sep/18 $${prove}\:{that}\:\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2cos}\:\mathrm{8}\theta}}} \\ $$$$\mathrm{2cos}\:\theta \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Sep/18 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{8}\theta}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{8}\theta\right)}\:}}…
Question Number 174556 by mnjuly1970 last updated on 04/Aug/22 Commented by infinityaction last updated on 04/Aug/22 $$\:\:\:\:\:\:\:\mathrm{tan}\alpha\:+\mathrm{3}\:=\:\mathrm{2sec}\alpha \\ $$$$\:\:\:\:\:\:\mathrm{3tan}^{\mathrm{2}} \alpha\:−\mathrm{6tan}\alpha\:−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{tan}\alpha\:\:=\:\:\:\frac{\mathrm{6}\:\pm\:\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{6}}\:\:\:=\:\:\frac{\mathrm{sin}\alpha\:}{\mathrm{cos}\alpha\:}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{sin}\alpha−\mathrm{cos}\alpha\:\:}{\mathrm{sin}\alpha+\mathrm{cos}\alpha\:\:}\:\:=\:\:\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{12}+\mathrm{4}\sqrt{\mathrm{6}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \\…
Question Number 43467 by peter frank last updated on 11/Sep/18 $${prove}\:{th}\mathrm{at}\:\mathrm{tan}\theta+\mathrm{2}{tan}\mathrm{2}\theta+\mathrm{4}{tan}\:\:\:\mathrm{4}\theta \\ $$$$+\mathrm{8cot8}\theta={cot}\theta \\ $$ Answered by ajfour last updated on 11/Sep/18 $$\mathrm{8cot}\:\mathrm{8}\theta+\mathrm{4tan}\:\mathrm{4}\theta+\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:\:=\:\frac{\mathrm{8}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}}…
Question Number 174435 by cortano1 last updated on 01/Aug/22 $$\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{3}°+…+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}°=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 43346 by peter frank last updated on 10/Sep/18 $$\mathrm{sin}\:{x}−\mathrm{sin}\:\mathrm{5}{x}=\mathrm{sin}\:\mathrm{3}{x}\:{find}\:{the}\:{angle} \\ $$$${that}\:{satisfied}\:{the}\:{equestion} \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 $${sinx}={sin}\mathrm{5}{x}+{sin}\mathrm{3}{x}…
Question Number 174346 by cortano1 last updated on 30/Jul/22 $$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°}\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}°}\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}°}\:−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{45}°\:}=? \\ $$ Answered by behi834171 last updated on 30/Jul/22 $$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{10}}=\mathrm{1}+{tg}^{\mathrm{2}}…