Question Number 85489 by oustmuchiya@gmail.com last updated on 22/Mar/20 $${Find}\:{all}\:{angles}\:{between}\:\mathrm{0}°\:{and}\:\mathrm{360}°,\:{for}\:{which}\:\mathrm{8}{sin}\theta=\mathrm{3}{cos}^{\mathrm{2}} \theta \\ $$ Commented by Tony Lin last updated on 22/Mar/20 $$\mathrm{8}{sin}\theta=\mathrm{3}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{3}{sin}^{\mathrm{2}}…
Question Number 85484 by oustmuchiya@gmail.com last updated on 22/Mar/20 $${show}\:{that}\:\frac{\mathrm{1}}{{sec}\theta+\mathrm{1}}+\frac{\mathrm{1}}{{sec}\theta−\mathrm{1}}\equiv\mathrm{2}{cosec}\theta{cot}\theta \\ $$ Answered by som(math1967) last updated on 18/Apr/20 $$\frac{\mathrm{sec}\:\theta−\mathrm{1}+{sec}\theta+\mathrm{1}}{\left({sec}\theta+\mathrm{1}\right)\left({sec}\theta−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}{sec}\theta}{{sec}^{\mathrm{2}} \theta−\mathrm{1}}=\frac{\mathrm{2}{sec}\theta}{{tan}^{\mathrm{2}} \theta}=\frac{\mathrm{2}{sec}\theta{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{2}}…
Question Number 85442 by Power last updated on 22/Mar/20 Commented by Power last updated on 22/Mar/20 $$\left[\boldsymbol{\mathrm{x}}\right]−\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{part}}=\boldsymbol{\mathrm{trunc}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\left\{\boldsymbol{\mathrm{x}}\right\}−\boldsymbol{\mathrm{fractional}}\:\boldsymbol{\mathrm{part}}=\boldsymbol{\mathrm{frac}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$ Answered by mr W…
Question Number 85386 by john santu last updated on 21/Mar/20 $$\mathrm{If}\:\mathrm{cot}\:{x}−\mathrm{cos}\:{x}\:=\:{n}\:{and}\: \\ $$$${m}^{\mathrm{2}} −{n}^{\mathrm{2}\:} =\:\mathrm{4}\sqrt{{mn}} \\ $$$${prove}\:{that}\:{m}\:=\:\mathrm{cot}\:{x}\:+\:\mathrm{cos}\:{x}\: \\ $$ Answered by mind is power last…
Question Number 85372 by john santu last updated on 21/Mar/20 $$\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{4cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{3}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$${find}\:\mathrm{tan}\:{x}.\: \\ $$ Answered by john santu last updated on 21/Mar/20…
Question Number 85349 by john santu last updated on 21/Mar/20 $$\sqrt{\mathrm{cos}\:\mathrm{2x}}\:+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}\:=\:\mathrm{2}\sqrt{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\: \\ $$$$\mathrm{x}\:\in\:\left(\mathrm{0},\mathrm{2}\pi\:\right) \\ $$ Commented by john santu last updated on 21/Mar/20 $$\sqrt{\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 85313 by john santu last updated on 20/Mar/20 $$\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{20}^{\mathrm{o}} \right)\:+\:\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{40}^{\mathrm{o}} \right)+\:\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{80}^{\mathrm{o}} \right)\:= \\ $$ Answered by john santu last updated…
Question Number 19760 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}\:+\:\mathrm{tan}\:{x}\:+\:\mathrm{cosec}\:{x}\:+ \\ $$$$\mathrm{cot}\:{x}\:+\:\mathrm{sec}\:{x}\:=\:\mathrm{7},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}. \\ $$ Answered by ajfour last updated on 15/Aug/17 $$\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}} \\…
Question Number 150761 by DELETED last updated on 15/Aug/21 Answered by DELETED last updated on 15/Aug/21 $$\left.\mathrm{1}.\mathrm{a}\right).\:\left(\mathrm{NH}_{\mathrm{4}} \right)_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \rightarrow\mathrm{N}_{\mathrm{2}} +\mathrm{4H}_{\mathrm{2}} \mathrm{O}+\mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{3}} \\…
Question Number 85146 by john santu last updated on 19/Mar/20 $$\mathrm{find}\:\mathrm{minimum}\:\&\:\mathrm{maximum}\:\mathrm{value}\: \\ $$$$\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:,\:−\pi\leqslant\mathrm{x}\leqslant\pi \\ $$ Commented by mr W last updated on…