Question Number 198423 by pascal889 last updated on 19/Oct/23 $$\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}+\mathrm{sin}\varnothing}{\mathrm{cos}\varnothing}\:=\frac{\mathrm{cos}\varnothing}{\mathrm{1}−\mathrm{sin}\varnothing} \\ $$$$\mathrm{Pls}\:\mathrm{help} \\ $$$$ \\ $$ Answered by BaliramKumar last updated on 19/Oct/23…
Question Number 198357 by pticantor last updated on 18/Oct/23 $$\boldsymbol{{let}}\:\:\boldsymbol{{a}}\in\mathbb{R},\:\boldsymbol{{z}}\in\mathbb{C} \\ $$$$\boldsymbol{{resolve}}\: \\ $$$$\left(\boldsymbol{{z}}+\mathrm{1}\right)^{\boldsymbol{{n}}} =\boldsymbol{{e}}^{\boldsymbol{{i}}\pi\boldsymbol{{na}}} \\ $$$$\boldsymbol{{deduce}}\:\boldsymbol{{that}}\:\boldsymbol{{P}}_{\boldsymbol{{n}}} =\underset{{k}=\mathrm{0}} {\overset{\boldsymbol{{n}}−\mathrm{1}} {\prod}}\boldsymbol{{sin}}\left(\boldsymbol{{a}}+\frac{\boldsymbol{{k}}\pi}{\boldsymbol{{n}}}\right) \\ $$ Terms of Service…
Question Number 197937 by mathlove last updated on 05/Oct/23 $$\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} \theta\centerdot{tan}\beta+{cos}^{\mathrm{2}} \beta=? \\ $$ Answered by Frix last updated on 05/Oct/23 $$\mathrm{It}\:\mathrm{is}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is},\:\mathrm{not}\:\mathrm{getting}\:\mathrm{better}\:\mathrm{any}\:\mathrm{way}. \\ $$$${p}=\mathrm{tan}\:\theta\:\wedge{q}=\mathrm{tan}\:\beta\:\rightarrow\:\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} {q}}{{p}^{\mathrm{2}}…
Question Number 197914 by a.lgnaoui last updated on 04/Oct/23 $$\boldsymbol{\mathrm{Determiner}}\:\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{hachuree}} \\ $$$$\left(\boldsymbol{\mathrm{voir}}\:\:\boldsymbol{\mathrm{figure}}\right) \\ $$$$\:\boldsymbol{\mathrm{BC}}=\mathrm{10}\boldsymbol{\mathrm{cm}}\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{B}}=\mathrm{45}°\:\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{30}° \\ $$ Commented by a.lgnaoui last updated on 04/Oct/23 Commented by…
Question Number 197908 by mathlove last updated on 03/Oct/23 $${sin}\mathrm{18}^{°} =? \\ $$ Answered by universe last updated on 03/Oct/23 $$\mathrm{sin}\:\left(\mathrm{36}°+\mathrm{54}°\right)\:=\:\mathrm{sin}\:\mathrm{90}° \\ $$$$\mathrm{36}°\:+\:\mathrm{54}°\:=\:\mathrm{90}°\: \\ $$$${x}\:=\:\mathrm{18}°\:…
Question Number 197881 by Abdullahrussell last updated on 02/Oct/23 Commented by Frix last updated on 02/Oct/23 $$\mathrm{2}−\sqrt{\mathrm{3}} \\ $$ Answered by Sutrisno last updated on…
Question Number 197848 by cortano12 last updated on 01/Oct/23 Answered by mr W last updated on 01/Oct/23 $${f}\left({x}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}}+\frac{{b}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:{x}}\geqslant\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:{x}}=\left({a}+{b}\right)^{\mathrm{2}}…
Question Number 197850 by tri26112004 last updated on 01/Oct/23 $${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{2}{sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{cos}\:\mathrm{2}{x}\right) \\ $$ Answered by Frix last updated on 01/Oct/23 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$−\mathrm{2sin}^{\mathrm{3}} \:{x}\:−\mathrm{sin}^{\mathrm{2}}…
Question Number 197795 by cortano12 last updated on 29/Sep/23 Answered by AST last updated on 29/Sep/23 $${Let}\:{y}={px};{x}\left({p}+\mathrm{1}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+{p}^{\mathrm{2}} \right)\:;{x}\neq\mathrm{0}\Rightarrow{p}^{\mathrm{2}} ={p}\Rightarrow{p}=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$${p}=\mathrm{0}\Rightarrow{y}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} ={x}\Rightarrow{x}=\mathrm{1} \\ $$$${p}=\mathrm{1}\Rightarrow\mathrm{2}{x}=\mathrm{2}{x}^{\mathrm{2}}…
Question Number 197717 by a.lgnaoui last updated on 27/Sep/23 $$\measuredangle\boldsymbol{\mathrm{A}}=\mathrm{62}\:\:\:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{43} \\ $$$$\boldsymbol{\mathrm{Determiner}}:\:\:\mathrm{a}=\measuredangle\boldsymbol{\mathrm{B}}\:\:\:\:\mathrm{c}=\measuredangle\boldsymbol{\mathrm{D}}\:\:\:\:\:\mathrm{b}=\measuredangle\boldsymbol{\mathrm{F}} \\ $$ Commented by a.lgnaoui last updated on 27/Sep/23 $$\measuredangle{EDF}=\mathrm{62}\:? \\ $$ Commented…