Question Number 8287 by lepan last updated on 06/Oct/16 $${Show}\:{that}\:{tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}. \\ $$ Answered by ridwan balatif last updated on 06/Oct/16 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{sin}\left(\alpha+\beta\right)}{\mathrm{cos}\left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha}{\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}{\left(\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}…
Question Number 73817 by Rio Michael last updated on 16/Nov/19 $${find}\:{the}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$${in}\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$$$\:\:{sin}\mathrm{2}\theta\:=\:{sec}\theta \\ $$ Answered by mr W last updated on 16/Nov/19…
Question Number 8273 by lepan last updated on 05/Oct/16 $${Express}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha\:{in}\:{the}\:{form}\: \\ $$$${Rsin}\left(\alpha+\beta\right)\:{where}\:{R}>\mathrm{0}\:{and}\:\mathrm{0}°<\beta<\mathrm{90}°. \\ $$$${Hence}\:{solve}\:{the}\:{equation}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha=\mathrm{2} \\ $$$${for}\:\mathrm{0}°<\alpha<\mathrm{270}°. \\ $$ Answered by prakash jain last updated on…
Question Number 8275 by lepan last updated on 05/Oct/16 $${Show}\:{that}\:{the}\:{followings} \\ $$$$\left({i}\right){sin}\left({a}+{b}\right)={sina}\:{cosb}\:+{cosa}\:{sinb} \\ $$$$\left({ii}\right){cos}\left({a}−{b}\right)={cosa}\:{cosb}\:+{sina}\:{sinb} \\ $$$$ \\ $$ Commented by sou1618 last updated on 05/Oct/16…
Question Number 8269 by tawakalitu last updated on 04/Oct/16 Commented by tawakalitu last updated on 04/Oct/16 $$\mathrm{Find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}.\:\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}. \\ $$ Answered by prakash jain last updated…
Question Number 8267 by lepan last updated on 04/Oct/16 $${Show}\:{that}\:{sinA}+{sinB}=\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}\:{cos}\frac{{A}−{B}}{\mathrm{2}}. \\ $$$$ \\ $$ Answered by Yozzias last updated on 04/Oct/16 $$\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{and}\:\mathrm{q}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}−\mathrm{B}\right). \\ $$$$\therefore\:\mathrm{2p}=\mathrm{A}+\mathrm{B}\:\mathrm{and}\:\mathrm{2q}=\mathrm{A}−\mathrm{B}. \\…
Question Number 8257 by lepan last updated on 04/Oct/16 $${If}\:{A}+{B}+{C}=\mathrm{90}°\:,{show}\:{that}\: \\ $$$${tanA}\:{tanB}+{tanB}\:{tanC}+{tanC}\:{tanA}=\mathrm{1}. \\ $$ Answered by sandy_suhendra last updated on 04/Oct/16 $$=\mathrm{tanA}\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\left(\mathrm{tanA}+\mathrm{tanB}\right) \\ $$$$=\mathrm{tanA}\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)\left[\mathrm{1}−\mathrm{tanA}\:\mathrm{tanB}\right] \\…
Question Number 8224 by trapti rathaur@ gmail.com last updated on 03/Oct/16 $${if}\:\:\varnothing\:{lies}\:{between}\:\:\:−\frac{\pi}{\mathrm{4}}\:{and}\:\:\frac{\pi}{\mathrm{4}}\:\:\:{then}\:\:{prove}\:{that} \\ $$$$\varnothing^{\mathrm{2}} =\mathrm{tan}\:^{\mathrm{2}} \varnothing\:−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\frac{\mathrm{tan}\:^{\mathrm{4}} \varnothing}{\mathrm{2}}\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}\right)\frac{\mathrm{tan}\:^{\mathrm{6}} \varnothing}{\mathrm{3}}\:+−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−{to}\:\infty\:{terms} \\ $$$$ \\ $$ Commented…
Question Number 8176 by lepan last updated on 02/Oct/16 $${Prove}\:{that}\:{cos}\mathrm{2}\theta=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}.{Hence}\:{deduce}\:{that}\: \\ $$$${tan}\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$ Answered by prakash jain last…
Question Number 73608 by mathocean1 last updated on 13/Nov/19 $$\:\mathrm{determiner} \\ $$$$\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{exacte}\:\mathrm{de}\:\mathrm{sin}\left(\frac{\mathrm{85}\Pi}{\mathrm{4}}\right) \\ $$$$ \\ $$ Commented by mathmax by abdo last updated on 13/Nov/19…