Question Number 15328 by tawa tawa last updated on 09/Jun/17 $$\mathrm{Prove}\:\mathrm{that}. \\ $$$$\mathrm{sec}^{\mathrm{4}} \left(\mathrm{x}\right)\:−\:\mathrm{cosec}^{\mathrm{4}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\:−\:\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{sec}^{\mathrm{4}} \left(\mathrm{x}\right)} \\ $$ Answered by RasheedSoomro last updated…
Question Number 15300 by Tinkutara last updated on 09/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\beta,\:\mathrm{0}\:<\:\alpha,\:\beta\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\left(\alpha\:+\:\beta\right)\:=\:−\frac{\mathrm{1}}{\mathrm{8}}\:. \\ $$ Commented by mrW1 last updated on 09/Jun/17 $$\alpha=\beta=\frac{\pi}{\mathrm{3}} \\…
Question Number 15301 by Tinkutara last updated on 09/Jun/17 $$\mathrm{With}\:\mathrm{the}\:\mathrm{help}\:\mathrm{of}\:\mathrm{graph},\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequation}\:\mathrm{tan}\:{x}\:>\:−\sqrt{\mathrm{3}}\:. \\ $$ Answered by mrW1 last updated on 09/Jun/17 $$\mathrm{x}\in\left(\mathrm{n}\pi−\frac{\pi}{\mathrm{3}},\:\mathrm{n}\pi+\frac{\pi}{\mathrm{2}}\right)\:\wedge\:\mathrm{n}\in\mathbb{Z} \\ $$ Commented…
Question Number 15298 by Tinkutara last updated on 09/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{and}\:{y} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:\mathrm{sin}\left({xy}\right)\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$ Commented by prakash jain last updated on 10/Jun/17 $${xy}={u} \\…
Question Number 15267 by Tinkutara last updated on 09/Jun/17 $$\mathrm{In}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{angled}\:\Delta{ABC},\:\mathrm{the} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}\:\mathrm{tan}\:{C}\:\mathrm{is}? \\ $$ Commented by prakash jain last updated on 09/Jun/17 $${A}=\mathrm{0}.\mathrm{1}°,{B}=\mathrm{89}.\mathrm{5}°,{C}=\mathrm{89}.\mathrm{5}° \\ $$$$\mathrm{tan}\:{A}\mathrm{tan}\:{B}\mathrm{tan}\:{C}=.\mathrm{518}…
Question Number 15264 by Tinkutara last updated on 08/Jun/17 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequation} \\ $$$$\mathrm{cos}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\geqslant\:\mathrm{0}\:\mathrm{is}\:\left[−\pi,\:\pi\right] \\ $$$$\left(\mathrm{1}\right)\:\left[\mathrm{0},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\left(\mathrm{2}\right)\:\left[−\frac{\mathrm{2}\pi}{\mathrm{3}},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\left(\mathrm{3}\right)\:\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\left(\mathrm{4}\right)\:\left[−\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right] \\ $$ Answered by mrW1…
Question Number 15263 by Tinkutara last updated on 08/Jun/17 $$\mathrm{The}\:\mathrm{equation}\:{a}\mathrm{sin}{x}\:+\:\mathrm{cos2}{x}\:=\:\mathrm{2}{a}\:−\:\mathrm{7} \\ $$$$\mathrm{possesses}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:>\:\mathrm{6} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}\:\leqslant\:{a}\:\leqslant\:\mathrm{6} \\ $$$$\left(\mathrm{3}\right)\:{a}\:>\:\mathrm{2} \\ $$$$\left(\mathrm{4}\right)\:{a} \\ $$ Answered by ajfour…
Question Number 80739 by jagoll last updated on 05/Feb/20 $$\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{2sin}\:^{\mathrm{2}} \theta=\mathrm{3}\bar {\:}\theta\in\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$$${what}\:{is}\:\theta\:? \\ $$ Commented by mr W last updated on 06/Feb/20…
Question Number 80595 by Power last updated on 04/Feb/20 Commented by Power last updated on 04/Feb/20 $$\mathrm{prove}\:\mathrm{that} \\ $$ Commented by Power last updated on…
Question Number 14991 by tawa tawa last updated on 06/Jun/17 Answered by ajfour last updated on 06/Jun/17 $$\:\:\:\:\:=\frac{\mathrm{tan}\:\theta}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right)}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{1}−\mathrm{tan}\:\theta\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:\theta−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{tan}\:\theta−\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{tan}\:^{\mathrm{3}} \theta−\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{tan}\:\theta−\mathrm{1}\right)}\: \\…