Question Number 161664 by cortano last updated on 21/Dec/21 $$\:\:\mathrm{5sec}\:\alpha\:−\mathrm{4}\:\mathrm{tan}\:\alpha\:=\:\mathrm{3cosec}\:\alpha \\ $$$$\:\:\frac{\mathrm{3cot}\:\alpha}{\mathrm{5}\:\mathrm{tan}\:\alpha−\mathrm{4}\:\mathrm{sec}\:\alpha}\:=?\: \\ $$ Answered by som(math1967) last updated on 21/Dec/21 $$\left(\mathrm{5}{sec}\alpha−\mathrm{4}{tan}\alpha\right)^{\mathrm{2}} =\mathrm{9}{cosec}^{\mathrm{2}} \alpha \\…
Question Number 30528 by abdo imad last updated on 22/Feb/18 $${simplify}\:\:{A}=\:{arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right)\:. \\ $$ Commented by abdo imad last updated on 23/Feb/18 $${we}\:{have}\:{A}={arctan}\left(\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$${A}={arctan}\left({cotan}\left(\frac{{x}}{\mathrm{2}}\right)\right)={arctan}\left(\:\frac{\mathrm{1}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\right)…
Question Number 30455 by ajfour last updated on 22/Feb/18 Commented by mrW2 last updated on 25/Feb/18 $${p}=\frac{{ac}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}}…
Question Number 95903 by 675480065 last updated on 28/May/20 Commented by PRITHWISH SEN 2 last updated on 28/May/20 $$\because\:\mathrm{x}>\mathrm{0} \\ $$$$\therefore\:\mathrm{x}^{\mathrm{2}} >\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{x}^{\mathrm{2}} <\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}<\mathrm{1}…
Question Number 161396 by abdullah_ff last updated on 17/Dec/21 $$\mathrm{If}\:{y}\:=\:{cos}^{\mathrm{2}} {x}\:+\:{sin}^{\mathrm{4}} {x}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\leq\:{y}\:\leq\:\mathrm{1} \\ $$ Commented by cortano last updated on 17/Dec/21…
Question Number 30235 by NECx last updated on 18/Feb/18 $${find}\:{the}\:{sum}\:{of}\:{the}\:{infinite} \\ $$$${series}\: \\ $$$$\:\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$ Commented by prof Abdo imad last updated…
Question Number 161254 by cortano last updated on 15/Dec/21 $$\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}=? \\ $$ Commented by bobhans last updated on 15/Dec/21 $$\mathcal{P}\:=\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)+\mathrm{1}}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)\left[\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)}{\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{14}}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)}\right]} \\ $$$$\:\mathcal{P}\:=\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)} \\ $$$$\:\mathcal{P}\:=\:\frac{−\mathrm{2}\left(\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right)+\mathrm{2cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)} \\…
Question Number 95703 by i jagooll last updated on 27/May/20 $$\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}^{\mathrm{o}} }\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:=? \\ $$ Commented by john santu last updated…
Question Number 161181 by cortano last updated on 13/Dec/21 $$\:\:{x}=\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:\theta}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:\theta}\right) \\ $$$$\:\mathrm{sin}\:{x}=? \\ $$ Commented by MJS_new last updated on 13/Dec/21 $$\mathrm{sin}\:\left(\mathrm{arctan}\:{a}\:−\mathrm{arctan}\:{b}\right)\:=\frac{{a}−{b}}{\:\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}}…
Question Number 30087 by rahul 19 last updated on 16/Feb/18 $$\mathrm{solve}: \\ $$$$\mathrm{cos3}{x}.{cos}^{\mathrm{3}} {x}+\mathrm{sin}\:\mathrm{3}{x}.\mathrm{sin}\:^{\mathrm{3}} {x}=\mathrm{0}. \\ $$ Answered by MJS last updated on 16/Feb/18 $${x}=\frac{\pi}{\mathrm{4}}+\frac{\pi\centerdot{z}}{\mathrm{2}};\:{z}\in\mathbb{Z}…