Question Number 18209 by b.e.h.i.8.3.417@gmail.com last updated on 16/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 16/Jul/17 $${excuse}\:{me}\:{for}\:{reposting}\:{this}. \\ $$ Answered by b.e.h.i.8.3.417@gmail.com last updated on…
Question Number 83739 by jagoll last updated on 05/Mar/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\left(\mathrm{sin}\:\mathrm{x}+\:\mathrm{csc}\:\mathrm{x}\:\right)^{\mathrm{2}} +\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x} \\ $$ Commented by MJS last updated on 05/Mar/20 $$−\infty \\ $$$$\mathrm{local}\:\mathrm{minima}\:\mathrm{at}…
Question Number 18185 by Tinkutara last updated on 16/Jul/17 $$\mathrm{Is}\:\mathrm{this}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false}? \\ $$$${a}\mathrm{cos}{A}\:+\:{b}\mathrm{cos}{B}\:+\:{c}\mathrm{cos}{C}\:=\:\frac{{abc}}{\mathrm{2}{R}^{\mathrm{2}} } \\ $$ Commented by b.e.h.i.8.3.417@gmail.com last updated on 16/Jul/17 $${yes}.\:{it}\:{is}\:{true}. \\ $$$${in}\:{triangle}\:{ABC},{draw}\:{altitudes}\:{of}…
Question Number 149259 by john_santu last updated on 04/Aug/21 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{sin}\:\left(\mathrm{41}°\right)\mathrm{sin}\:\left(\mathrm{19}°\right)+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)=? \\ $$ Answered by john_santu last updated on 04/Aug/21 $$\Leftrightarrow\mathrm{sin}\:\mathrm{41}°\mathrm{sin}\:\mathrm{19}°=\frac{−\mathrm{2sin}\:\mathrm{4l}°\mathrm{sin}\:\mathrm{19}°}{−\mathrm{2}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{60}°−\mathrm{cos}\:\mathrm{22}°}{−\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{4}}…
Question Number 18184 by Tinkutara last updated on 16/Jul/17 $$\left({m}+\mathrm{2}\right)\mathrm{sin}\theta\:+\:\left(\mathrm{2}{m}−\mathrm{1}\right)\mathrm{cos}\theta\:=\:\mathrm{2}{m}+\mathrm{1},\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:+\:\mathrm{1}} \\ $$ Commented by prakash…
Question Number 83680 by Power last updated on 05/Mar/20 Commented by john santu last updated on 05/Mar/20 $$\mathrm{considering}\:\mathrm{A}+\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} \\ $$$$\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} −\mathrm{A}\:\Rightarrow\:\mathrm{tan}\:\mathrm{B}\:=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{tan}\:\mathrm{A}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\mathrm{A}} \\ $$$$\mathrm{tan}\:\mathrm{B}\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}} \\ $$$$\sqrt{\mathrm{3}}\:+\:\mathrm{tan}\:\mathrm{B}\:=\:\sqrt{\mathrm{3}}+\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}}…
Question Number 83654 by jagoll last updated on 05/Mar/20 $$\mathrm{solve}\:\mathrm{this}\:\mathrm{equation}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{4}} {x}=\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{cos}\:^{\mathrm{4}} {x} \\ $$ Commented by jagoll last updated on 05/Mar/20…
Question Number 18114 by ibraheem160 last updated on 15/Jul/17 $$\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}{\left.\mathrm{1}−\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}} \\ $$$$\Rightarrow\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}} \\ $$$$\mathrm{recall}\:\mathrm{that}:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{1}=\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\theta}+\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} \\ $$$$=\mathrm{cosec}\theta+\mathrm{cot}\theta…
Question Number 18093 by Tinkutara last updated on 15/Jul/17 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{sin}{x}\:+\:\mathrm{sin2}{x}\:+\:\mathrm{2sin}{x}\mathrm{sin2}{x} \\ $$$$=\:\mathrm{2cos}{x}\:+\:\mathrm{cos2}{x}\:\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by}\:\mathrm{values} \\ $$$$\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{6}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\pi}{\mathrm{2}}\:,\:{n}\:\in\:{I} \\ $$…
Question Number 18091 by Tinkutara last updated on 15/Jul/17 $$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{statement}\left(\mathrm{s}\right) \\ $$$$\mathrm{is}/\mathrm{are}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{cos}\left(\mathrm{sin1}\right)\:>\:\mathrm{sin}\left(\mathrm{cos1}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}\left(\mathrm{sin1}.\mathrm{5}\right)\:>\:\mathrm{sin}\left(\mathrm{cos1}.\mathrm{5}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{cos}\left(\mathrm{sin}\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:>\:\mathrm{sin}\left(\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{18}}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{cos}\left(\mathrm{sin}\frac{\mathrm{5}\pi}{\mathrm{18}}\right)\:>\:\mathrm{sin}\left(\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{18}}\right) \\ $$ Answered by Tinkutara…