Question Number 107978 by anonymous last updated on 13/Aug/20 $${on}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\pi\right]\:{solve} \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{3}{secx}=\:−\mathrm{3} \\ $$$$ \\ $$ Answered by bemath last updated on 14/Aug/20 $$\:\:\:\frac{\bullet\mathscr{B}{emath}\bullet}{\mathscr{C}{ooll}}…
Question Number 107977 by anonymous last updated on 13/Aug/20 $${On}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{solve} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}\:+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$ \\ $$ Answered by bemath last updated on 13/Aug/20 $$\:\:\frac{\mathcal{B}{e}\frac{\mathcal{M}{ath}}{\heartsuit}}{{joss}} \\…
Question Number 107932 by bemath last updated on 13/Aug/20 $$\:\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet} \\ $$$${Given}\:\begin{cases}{\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{tan}\:{x}\:=\:\mathrm{2}\:}\end{cases} \\ $$$${find}\:\:\mathrm{tan}\:{y}\:? \\ $$ Commented by mnjuly1970 last updated on 13/Aug/20 $$\:{excellent}. \\…
Question Number 173447 by ajfour last updated on 11/Jul/22 $${Find}\:\mathrm{tan}\:\left(\mathrm{142}.\mathrm{5}°\right)\:{without} \\ $$$${tables}. \\ $$ Commented by mr W last updated on 12/Jul/22 $$\boldsymbol{{welcome}}\:\boldsymbol{{back}}\:\boldsymbol{{sir}}! \\ $$$$…
Question Number 107900 by bemath last updated on 13/Aug/20 $$\:\:\:\:\:\:\:\frac{\Sigma\:\mathcal{B}{e}\mathcal{M}{ath}\:\Sigma}{\Box} \\ $$$$\:\:{Given}\:\mathrm{tan}\:{x}−\mathrm{sec}\:{x}\:=\:\vartheta\: \\ $$$$\:{then}\:\mathrm{sin}\:{x}\:=\:? \\ $$ Answered by bemath last updated on 13/Aug/20 $$\frac{\mathrm{sin}\:{x}−\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\upsilon\:\Rightarrow\:\mathrm{sin}\:{x}−\mathrm{1}=\upsilon\:\mathrm{cos}\:{x}\:…\left(\mathrm{1}\right) \\…
Question Number 42175 by lucha116 last updated on 19/Aug/18 $${sin}\mathrm{3}{x}+{sin}\mathrm{5}{x}=\mathrm{2}\left({cos}^{\mathrm{2}} \mathrm{2}{x}−{sin}^{\mathrm{2}} \mathrm{3}{x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42174 by lucha116 last updated on 19/Aug/18 $${cos}^{\mathrm{2}} \mathrm{3}{xcos}\mathrm{2}{x}−{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18 $$\left(\frac{\mathrm{1}+{cos}\mathrm{6}{x}}{\mathrm{2}}\right){cos}\mathrm{2}{x}−\left(\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${cos}\mathrm{2}{x}+{cos}\mathrm{2}{x}.{cos}\mathrm{6}{x}−\mathrm{1}−{cos}\mathrm{2}{x}=\mathrm{0} \\…
Question Number 107684 by bemath last updated on 12/Aug/20 $$\:\:\:\:\:“\mathcal{B}{e}\mathcal{M}{ath}“ \\ $$$$\:\mathrm{ln}\:\mathrm{tan}\:\mathrm{1}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{2}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{ln}\:\mathrm{tan}\:\mathrm{89}°=? \\ $$ Answered by Her_Majesty last updated on 12/Aug/20 $${ln}\:{tan}\:{x}°\:=−{ln}\:{tan}\:\left(\mathrm{90}−{x}\right)° \\ $$$$\Rightarrow\:{answer}\:{is}\:{ln}\:{tan}\:\mathrm{45}°\:=\mathrm{0} \\…
Question Number 107672 by bemath last updated on 12/Aug/20 $$\:\:\:\:\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\spadesuit \\ $$$$\:\:\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:? \\ $$ Commented by Her_Majesty last updated on 12/Aug/20 $$\mathrm{2} \\ $$ Commented…
Question Number 107555 by anonymous last updated on 11/Aug/20 Answered by bshahid010 last updated on 11/Aug/20 $${if}\:{cos}\alpha=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{sin}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow{cos}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${Now}\:{cos}\left(\alpha+\beta\right)={cos}\alpha.{cos}\beta−{sin}\alpha.{sin}\beta \\ $$$$\Rightarrow−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow−\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)…