Question Number 195263 by dimentri last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{{y}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\mathrm{2}}\\{\mathrm{4sin}\:{x}−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=?\: \\ $$ Answered by cortano12 last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{\:\cancel{\underline{\underbrace{ }}}}\end{cases} \\ $$…
Question Number 195097 by Rupesh123 last updated on 24/Jul/23 Answered by Rasheed.Sindhi last updated on 24/Jul/23 $$\mathrm{49}^{\mathrm{sin}\:{x}} =\sqrt[{\mathrm{cos}\:{x}}]{\mathrm{7}}\: \\ $$$$\mathrm{7}^{\mathrm{2}\:\mathrm{sin}\:{x}} =\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{cos}\:{x}}} \\ $$$$\mathrm{2}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{cos}\:{x}} \\ $$$$\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}=\mathrm{1}…
Question Number 194937 by dimentri last updated on 20/Jul/23 $$\:\:{If}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{24}}\right)=\:\left(\sqrt{{a}}−\sqrt{{b}}\right)\left(\sqrt{{c}}−\sqrt{{d}}\right) \\ $$$$\:\:{where}\:{a},{b},{c},{d}\:{are}\:{postive}\:{numbers}. \\ $$$$\:\:{Find}\:{the}\:{value}\:{of}\:\left({a}+{b}+{c}+{d}+\mathrm{2}\right) \\ $$ Answered by BaliramKumar last updated on 20/Jul/23 $$\mathrm{tan}\left(\frac{\pi}{\mathrm{24}}\right)\:=\:\mathrm{tan7}.\mathrm{5}° \\…
Question Number 194903 by cortano12 last updated on 19/Jul/23 Commented by cortano12 last updated on 19/Jul/23 $$\:\:\:{x}\:=\:\sqrt{\mathrm{39}}\:\:\left(×\right) \\ $$$$\:\:\:{x}\:=\:\sqrt{\mathrm{30}}\:\left(\checkmark\right) \\ $$ Answered by MM42 last…
Question Number 194913 by cortano12 last updated on 19/Jul/23 Answered by witcher3 last updated on 19/Jul/23 $$\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}''\left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\left.\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \mathrm{f}''\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}'\mathrm{cos}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int\mathrm{f}'\mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\left.=\left.−\mathrm{f}'\left(\mathrm{0}\right)+\mathrm{fsin}\left(\mathrm{x}\right)\right]_{\mathrm{0}}…
Question Number 194881 by tri26112004 last updated on 18/Jul/23 $${Give}\:\bigtriangleup{ABC}\: \\ $$$${Proof}:\:{sin}\:{A}\:+\:{sin}\:{B}\:+\:{sin}\:{C}\:>\:\mathrm{2} \\ $$ Answered by Frix last updated on 19/Jul/23 $$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}. \\ $$$$\mathrm{0}<\mathrm{sin}\:{A}\:+\mathrm{sin}\:{B}\:+\mathrm{sin}\:{C}\:\:\leqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 194826 by cortano12 last updated on 16/Jul/23 $$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{19}°\:=\:{p}\: \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{7}°\:=? \\ $$ Answered by dimentri last updated on 16/Jul/23 $$\:\:\:\mathrm{tan}\:\mathrm{38}°\:=\:\mathrm{tan}\:\left(\mathrm{45}°−\mathrm{7}\right) \\ $$$$\:\:\:\frac{\mathrm{2tan}\:\mathrm{19}°}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{19}°}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{7}°}…
Question Number 194767 by horsebrand11 last updated on 15/Jul/23 $$\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2}\: \\ $$$$\:\:\:\frac{\mathrm{8sin}\:\theta+\mathrm{5cos}\:\theta}{\mathrm{sin}\:^{\mathrm{3}} \theta+\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{cos}\:\theta}\:=? \\ $$ Answered by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{sec}\:^{\mathrm{2}} {x}=\mathrm{5}\:,\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 194766 by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{1}+\mathrm{2cot}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\:\mathrm{3}\: \\ $$$$\:\:\:{x}=? \\ $$ Answered by Frix last updated on 15/Jul/23 $$\mathrm{1}+\mathrm{2cor}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\mathrm{3} \\ $$$$\mathrm{cot}^{\mathrm{2}} \:{x}\:=\mathrm{3}…
Question Number 194697 by cortano12 last updated on 13/Jul/23 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by MM42 last updated on 13/Jul/23 $$\frac{{tanx}−{tan}\mathrm{3}{x}}{{tanx}}=\mathrm{3}\Rightarrow\frac{{tan}\mathrm{3}{x}}{{tanx}}=−\mathrm{2} \\ $$$$\frac{{cotx}}{{cotx}+{cot}\mathrm{3}{x}}=\frac{{tan}\mathrm{3}{x}}{{tanx}+{tan}\mathrm{3}{x}} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{2}}=\mathrm{2}\:\checkmark \\…