Question Number 128995 by bramlexs22 last updated on 12/Jan/21 $$\mathrm{If}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\:−\mathrm{3tan}\:\mathrm{x}=\mathrm{1}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mid\:\mathrm{cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:\mid.\: \\ $$ Commented by liberty last…
Question Number 63460 by Rio Michael last updated on 04/Jul/19 $${factorise} \\ $$$${cos}\theta−{cos}\mathrm{3}\theta−{cos}\mathrm{5}\theta\:+{cos}\mathrm{7}\theta \\ $$ Commented by Tony Lin last updated on 04/Jul/19 $${cos}\theta−{cos}\mathrm{3}\theta−\left({cos}\mathrm{5}\theta−{cos}\mathrm{7}\theta\right) \\…
Question Number 63430 by rajesh4661kumar@gamil.com last updated on 04/Jul/19 Answered by MJS last updated on 04/Jul/19 $${y}=\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{x}−\mathrm{1}} \\ $$$$\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant\sqrt{{x}−\mathrm{1}}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{y}\leqslant\frac{\pi}{\mathrm{2}} \\ $$ Terms…
Question Number 63426 by Rio Michael last updated on 04/Jul/19 $${show}\:{that}\: \\ $$$$\left({a}\right)\:{cos}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left({x}\right)\:+{sin}^{−\mathrm{1}} \left({x}\right)\right]=\:−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$$$\left({b}\right)\:\frac{{sin}\alpha\:+\:{sin}\beta}{{cos}\alpha−{cos}\beta}={cot}\left(\frac{\beta−\alpha}{\mathrm{2}}\right) \\ $$$$\left({c}\right)\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}+{p}\right)\approxeq\:\mathrm{1}−\sqrt{\mathrm{3}}\:{if}\:{p}\:{is}\:{small}\:{enough}\:{to}\:{neglect}\:{p}^{\mathrm{2}} . \\ $$$$\left({d}\right)\:{if}\:\theta\:=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right),\:{show}\:{that}\:{sin}\theta−{cos}\theta\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 63423 by minh2001 last updated on 04/Jul/19 $${Prove}\:{that}\: \\ $$$$\frac{{sin}^{\mathrm{2}} \left(\mathrm{36}°\right)}{{cos}^{\mathrm{2}} \left(\mathrm{72}°\right)}+{sin}^{\mathrm{2}} \left(\mathrm{72}°\right){cos}\left(\mathrm{36}°\right)=\frac{\mathrm{45}+\mathrm{11}\sqrt{\mathrm{5}}}{\mathrm{16}} \\ $$ Commented by Tony Lin last updated on 04/Jul/19…
Question Number 63363 by rajesh4661kumar@gamil.com last updated on 03/Jul/19 Commented by Prithwish sen last updated on 03/Jul/19 $$\mathrm{tan70}=\frac{\mathrm{tan40}+\mathrm{tan20}+\mathrm{tan10}−\mathrm{tan40tan20tan10}}{\mathrm{1}−\mathrm{tan40tan20}−\mathrm{tan40tan10}−\mathrm{tan20tan10}} \\ $$$$\mathrm{tan70}\:−\:\mathrm{tan70tan40tan20}−\mathrm{tan70tan40tan10}−\mathrm{tan70tan20tan10} \\ $$$$=\:\mathrm{tan40}+\mathrm{tan20}+\mathrm{tan10}−\mathrm{tan40tan20tan10} \\ $$$$\because\mathrm{tan70tan20}=\mathrm{1}\:\therefore\mathrm{tan70tan40tan20}=\mathrm{tan40}\:\mathrm{and}\:\mathrm{tan70tan20tan10}\:=\:\mathrm{tan10} \\…
Question Number 63361 by rajesh4661kumar@gamil.com last updated on 03/Jul/19 Commented by Prithwish sen last updated on 03/Jul/19 $$\mathrm{tan3A}\:=\:\mathrm{tan}\left(\mathrm{2A}+\mathrm{A}\right)=\frac{\mathrm{tan2A}+\mathrm{tanA}}{\mathrm{1}−\mathrm{tan2AtanA}} \\ $$$$\mathrm{tan3A}−\mathrm{tan3Atan2AtanA}\:=\:\mathrm{tan2A}+\mathrm{tanA} \\ $$$$\mathrm{tan3A}−\mathrm{tan2A}−\mathrm{tanA}\:=\:\mathrm{tan3Atan2AtanA}\:\mathrm{proved}. \\ $$ Commented…
Question Number 63300 by Rio Michael last updated on 02/Jul/19 $${show}\:{that}\:\: \\ $$$$\left.{a}\right)\:\mathrm{1}\:+\:{tan}\:\left(\frac{\pi}{\mathrm{4}}\:+\:{A}\right)\:=\:\frac{\mathrm{2}}{\mathrm{1}−{tanA}} \\ $$$$\left.{b}\right)\:\mathrm{2}{cos}\mathrm{2}\theta{sin}\theta\:+\:\mathrm{9}{sin}\theta\:+\:\mathrm{3}\:\equiv\:\mathrm{11}{sin}\theta\:−\:\mathrm{4}{sin}^{\mathrm{3}} \theta\:+\:\mathrm{3} \\ $$ Commented by kaivan.ahmadi last updated on 02/Jul/19…
Question Number 128808 by liberty last updated on 10/Jan/21 $$\:\mathrm{If}\:{x}\:\mathrm{cos}\:{q}\:−\:\mathrm{sin}\:{q}\:=\:\mathrm{1}\:\mathrm{then}\:{x}^{\mathrm{2}} +\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{sin}\:{q}\:=?\: \\ $$ Commented by benjo_mathlover last updated on 10/Jan/21 $$\:{x}\:=\:\frac{\mathrm{cos}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}+\mathrm{sin}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}+\mathrm{2sin}\:\frac{{q}}{\mathrm{2}}\mathrm{cos}\:\frac{{q}}{\mathrm{2}}}{\mathrm{cos}\:^{\mathrm{2}} \frac{{q}}{\mathrm{2}}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 128788 by bramlexs22 last updated on 10/Jan/21 $$\:\mathrm{Prove}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{tan}\:\mathrm{x}\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{4x}\right)\:=\mathrm{2}? \\ $$ Commented by bramlexs22 last updated on 10/Jan/21 $$\mathrm{i}\:\mathrm{think}\:\mathrm{its}\:\mathrm{not}\:\mathrm{true}\:! \\ $$ Commented by mr…