Question Number 20721 by ajfour last updated on 01/Sep/17 $$\mathrm{tan}\:{x}\mathrm{tan}\:{z}=\mathrm{3} \\ $$$$\mathrm{tan}\:{y}\mathrm{tan}\:{z}=\mathrm{6} \\ $$$${x}+{y}+{z}\:=\:\pi \\ $$$${Solve}\:{for}\:{x},\:{y},\:{and}\:{z}. \\ $$ Answered by sma3l2996 last updated on 02/Sep/17…
Question Number 20703 by tammi last updated on 01/Sep/17 $$\frac{\mathrm{sin}\alpha+\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\:\alpha+\mathrm{cos}\:\mathrm{3}\alpha+\mathrm{cos}\:\mathrm{5}\alpha}=\mathrm{tan}\:\mathrm{3}\alpha \\ $$ Commented by myintkhaing last updated on 01/Sep/17 $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{\mathrm{2sin3}\alpha\mathrm{cos2}\alpha+\mathrm{sin3}\alpha}{\mathrm{2cos3}\alpha\mathrm{cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{sin3}\alpha}{\mathrm{cos3}\alpha}=\mathrm{tan3}\alpha \\ $$$$\mathrm{proved} \\ $$ Answered…
Question Number 20704 by tammi last updated on 01/Sep/17 $$\mathrm{4cos}\:\theta\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\theta\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{3}}+\theta\right)=\mathrm{cos}\:\mathrm{3}\theta \\ $$ Commented by myintkhaing last updated on 01/Sep/17 $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{2cos}\theta\left[\mathrm{cos}\left(\mathrm{2}\pi+\theta\right)+\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\theta\left[\mathrm{cos2}\theta−\mathrm{cos}\frac{\pi}{\mathrm{3}}\right]=\mathrm{2cos}\theta\left[\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4cos}^{\mathrm{3}}…
Question Number 20702 by tammi last updated on 01/Sep/17 $$\mathrm{tan}\:\mathrm{20}°\mathrm{tan}\:\mathrm{40}°\mathrm{tan}\:\mathrm{80}°=\sqrt{\mathrm{3}} \\ $$ Answered by Tinkutara last updated on 01/Sep/17 $$\mathrm{Using}\:\mathrm{tan}\:\theta\mathrm{tan}\:\left(\mathrm{60}−\theta\right)\mathrm{tan}\:\left(\mathrm{60}+\theta\right)=\mathrm{tan}\:\mathrm{3}\theta \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{20tan}\:\mathrm{40tan}\:\mathrm{80}=\mathrm{tan}\:\mathrm{60}=\sqrt{\mathrm{3}} \\ $$ Terms…
Question Number 20653 by Tinkutara last updated on 30/Aug/17 $$\mathrm{Length}\:\mathrm{of}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function} \\ $$$${f}\left(\theta\right)\:=\:\mathrm{cos}^{\mathrm{2}} \:\theta\:−\:\mathrm{6}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{2} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{10}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\sqrt{\mathrm{10}} \\…
Question Number 20647 by oyshi last updated on 30/Aug/17 $${prove}\:{it}, \\ $$$$\mathrm{tan}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)+\mathrm{tan}\:\left(\alpha−\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{4sin}\:\mathrm{2}\alpha}{\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \alpha} \\ $$ Answered by $@ty@m last updated on 30/Aug/17 $${LHS}=\frac{{tan}\alpha+{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}−{tan}\alpha.{tan}\frac{\pi}{\mathrm{3}}}+\frac{{tan}\alpha−{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}+{tan}\alpha.{tan}\frac{\pi}{\mathrm{3}}} \\ $$$$=\frac{{tan}\alpha+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}.{tan}\alpha}+\frac{{tan}\alpha−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}.{tan}\alpha}…
Question Number 20648 by Joel577 last updated on 30/Aug/17 $$\mathrm{If}\:\:\frac{{F}}{\mathrm{sin}\:\left({A}\:−\:{a}\right)}\:=\:\frac{{W}}{\mathrm{sin}\:{A}},\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{tan}\:{A}\:=\:\frac{{W}\:\mathrm{sin}\:{a}}{{W}\:\mathrm{cos}\:{a}\:−\:{F}} \\ $$ Answered by $@ty@m last updated on 30/Aug/17 $${Given}, \\ $$$$\:\frac{{F}}{\mathrm{sin}\:\left({A}\:−\:{a}\right)}\:=\:\frac{{W}}{\mathrm{sin}\:{A}} \\…
Question Number 20646 by oyshi last updated on 30/Aug/17 $${if}\:\mathrm{cos}\:\left(\beta−\gamma\right)+\mathrm{cos}\left(\:\gamma−\alpha\right)+\mathrm{cos}\:\left(\alpha−\beta\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${so}\:{proof}\:{it}, \\ $$$$\Sigma\mathrm{cos}\:\alpha=\mathrm{0},\Sigma\mathrm{sin}\:\alpha=\mathrm{0} \\ $$ Answered by $@ty@m last updated on 31/Aug/17 $${Given}, \\…
Question Number 86168 by jagoll last updated on 27/Mar/20 $$\mathrm{given}\: \\ $$$$\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{y}\:=\:\mathrm{2sin}\:\left(\mathrm{x}+\mathrm{y}\right)\: \\ $$$$\mathrm{x}+\mathrm{y}\:\neq\:\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\:\mathrm{tan}\:\frac{\mathrm{y}}{\mathrm{2}}\:=\: \\ $$ Answered by john santu last…
Question Number 20639 by oyshi last updated on 30/Aug/17 $${if}\:\mathrm{cos}\:\left({A}+{B}\right)\mathrm{sin}\:\left({C}+{D}\right)=\mathrm{cos}\:\left({A}−{B}\right) \\ $$$$\mathrm{sin}\:\left({C}−{D}\right) \\ $$$${so}\:{proof}\:\mathrm{cot}\:{A}\mathrm{cot}\:{B}\mathrm{cot}\:{C}=\mathrm{cot}\:{D} \\ $$ Answered by Tinkutara last updated on 30/Aug/17 $$\frac{\mathrm{sin}\:\left({C}+{D}\right)}{\mathrm{sin}\:\left({C}−{D}\right)}=\frac{\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{cos}\:\left({A}+{B}\right)} \\…