Question Number 105764 by bobhans last updated on 31/Jul/20 $$\left(\mathrm{1}\right){If}\:\mathrm{cos}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\:{and}\:\mathrm{sin}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{5}}{\mathrm{13}} \\ $$$${where}\:\mathrm{0}\:<\:\alpha<\:\frac{\pi}{\mathrm{4}}.\:{Find}\:\mathrm{tan}\:\mathrm{2}\alpha\:. \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)? \\ $$ Commented by Dwaipayan Shikari last…
Question Number 40231 by Raj Singh last updated on 17/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18 $$\frac{{cosA}−{sinA}+\mathrm{1}}{{cosA}+{sinA}−\mathrm{1}} \\ $$$${a}={sinA}\:\:\:{b}={cosA}\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{\left({b}−{a}+\mathrm{1}\right)\left({b}+{a}+\mathrm{1}\right)}{\left({b}+{a}−\mathrm{1}\right)\left({b}+{a}+\mathrm{1}\right)} \\…
Question Number 105766 by bramlex last updated on 31/Jul/20 $${If}\:{p}\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}\right)\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}\right)\mathrm{sin}\:\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right) \\ $$$${find}\:{the}\:{value}\:{of}\:{p}. \\ $$ Answered by som(math1967) last updated on 31/Jul/20 $$\mathrm{let}\:\frac{\pi}{\mathrm{18}}=\alpha \\ $$$$\therefore\mathrm{p}=\mathrm{sin}\alpha\mathrm{sin5}\alpha\mathrm{sin7}\alpha \\…
Question Number 105759 by bramlex last updated on 31/Jul/20 $${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{solution}\:{of} \\ $$$${equation}\:{a}\:\mathrm{tan}\:\theta\:+\:{b}\:\mathrm{sec}\:\theta\:=\:{c}\:.\: \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{tan}\:\left(\alpha+\beta\right). \\ $$ Commented by bramlex last updated on 31/Jul/20 $${thx}\:{both} \\…
Question Number 105692 by bramlex last updated on 31/Jul/20 $$\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{7}{x}\right)=\mathrm{1} \\ $$ Answered by john santu last updated on 31/Jul/20 $$\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{7}{x}\right)=\mathrm{1}…
Question Number 105650 by bemath last updated on 30/Jul/20 $$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)+\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$${find}\:{x} \\ $$ Answered by bramlex last updated on 30/Jul/20 $$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)=\frac{\pi}{\mathrm{6}}−\mathrm{cos}^{−\mathrm{1}}…
Question Number 105619 by bobhans last updated on 30/Jul/20 $${find}\:\mathcal{G}{eneral}\:{solution}\:\mathrm{cot}\:{x}+\mathrm{cot}\:\mathrm{2}{x}+\mathrm{cot3}{x}=\:\mathrm{0} \\ $$$$ \\ $$ Answered by bemath last updated on 30/Jul/20 $$\Leftrightarrow\mathrm{cot}\:\left({x}\right)+\frac{\mathrm{cot}\:^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{\mathrm{2cot}\:\left({x}\right)}+\frac{\mathrm{3cot}\:\left({x}\right)−\mathrm{cot}\:^{\mathrm{3}} \left({x}\right)}{\mathrm{1}−\mathrm{3cot}\:^{\mathrm{2}} \left({x}\right)}=\:\mathrm{0}…
Question Number 105575 by Algoritm last updated on 30/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170968 by 281981 last updated on 05/Jun/22 Commented by mr W last updated on 05/Jun/22 $$\mathrm{tan}^{\mathrm{2}} \:\left({x}+{y}\right)+\mathrm{cos}^{\mathrm{2}} \:\left({x}+{y}\right)−\mathrm{1}+\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\left({x}+{y}\right)−\mathrm{sin}^{\mathrm{2}} \:\left({x}+{y}\right)+\left({y}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 105366 by bemath last updated on 28/Jul/20 Answered by $@y@m last updated on 28/Jul/20 $$\mathrm{15}.\:\frac{\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}+\mathrm{1}\right)+\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}−\mathrm{1}\right)}{\mathrm{sec}^{\mathrm{2}} \:{A}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2tan}\:{A}\mathrm{sec}\:{A}}{\mathrm{tan}\:^{\mathrm{2}} {A}} \\ $$$$=\mathrm{2cosec}\:{A} \\ $$…