Question Number 111002 by bemath last updated on 01/Sep/20 $$\sqrt{\mathrm{bemath}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{14}°+\mathrm{cos}\:\mathrm{14}°\mathrm{tan}\:\mathrm{38}°−\mathrm{1}=? \\ $$ Commented by Khanacademy last updated on 01/Sep/20 $$\alpha\boldsymbol{{beMath\_uz}}\:\:\:\boldsymbol{{sizning}}\:\:\boldsymbol{{kanalingizmi}} \\ $$ Answered…
Question Number 45440 by Rio Michael last updated on 13/Oct/18 $${show}\:{that}\: \\ $$$$\frac{\mathrm{1}+\mathrm{2}{sin}\mathrm{2}\theta−{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta+{cos}\mathrm{2}\theta}\:\equiv\:{tan}\theta \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18 $$\frac{\mathrm{1}−{cos}\mathrm{2}\theta+{sin}\mathrm{2}\theta}{\mathrm{1}+{cos}\mathrm{2}\theta+{sin}\mathrm{2}\theta} \\ $$$$\frac{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 176490 by mnjuly1970 last updated on 20/Sep/22 $$ \\ $$$$\:\:\:\:\:{solve}\:\:\:\left(\:{x},{y}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\begin{cases}{\:\mathrm{tan}\left({x}\:\right)\:+\:\mathrm{tan}\:\left({y}\:\right)=\mathrm{2}}\\{\:\mathrm{tan}\left(\mathrm{2}{x}\:\right)\:+\:\mathrm{tan}\left(\:\mathrm{2}{y}\:\right)\:=\:\mathrm{2}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$ Answered by ajfour…
Question Number 176468 by blackmamba last updated on 20/Sep/22 $$\:\:\frac{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{18}°\right)}{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{12}°\right)}\:=\sqrt{\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{sin}\:\mathrm{48}°}}\: \\ $$$$\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\sqrt{\mathrm{tan}\:{M}}\:.\sqrt{\mathrm{tan}\:{N}} \\ $$$$\:\:\mathrm{0}°<{M},{N}<\mathrm{90}°\:\Rightarrow{M}+{N}=?° \\ $$ Commented by Peace last updated on 20/Sep/22 $${i}\:{started}\:{Withe}\:\mathrm{48}\:,{i}\:{see}\:{no}\:{nice}\:{Form} \\…
Question Number 176462 by blackmamba last updated on 19/Sep/22 $${In}\:\Delta{ABC}\:{given}\:\frac{\mathrm{2}{a}}{\mathrm{tan}\:{A}}\:=\:\frac{{b}}{\mathrm{tan}\:{B}}\: \\ $$$$\:{then}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{cos}\:^{\mathrm{2}} {B}}{\mathrm{cos}\:^{\mathrm{2}} {A}+\mathrm{cos}\:^{\mathrm{2}} {B}}=? \\ $$ Commented by cortano1 last updated on 19/Sep/22…
Question Number 45270 by ajfour last updated on 11/Oct/18 Commented by ajfour last updated on 11/Oct/18 $${Find}\:\boldsymbol{\alpha}\:{in}\:{terms}\:{of}\:{R}\:{and}\:{r}. \\ $$ Answered by MrW3 last updated on…
Question Number 176211 by adhigenz last updated on 15/Sep/22 Answered by mahdipoor last updated on 15/Sep/22 $$\mathrm{2021}\frac{{cos}\left({a}+{b}\right)}{{cosa}}{sinb}=\frac{{sina}}{{cosa}} \\ $$$$\mathrm{2021}\frac{{cosa}.{cosb}−{sina}.{sinb}}{{cosa}}{sinb}={tana} \\ $$$$\mathrm{2021}{cosb}.{sinb}−\mathrm{2021}{tana}.{sin}^{\mathrm{2}} {b}={tana} \\ $$$$\Rightarrow{tana}=\frac{\mathrm{2021}{cosb}.{sinb}}{\mathrm{1}+\mathrm{2021}{sin}^{\mathrm{2}} {b}}={f}\left({b}\right)…
Question Number 176164 by HeferH last updated on 14/Sep/22 $$\:{Proof}\:{that}\:: \\ $$$$\: \\ $$$$\:\sqrt{\frac{\mathrm{1}\:−\:\mathrm{cos}\:{x}}{\mathrm{1}\:+\:\mathrm{cos}\:{x}}}\:+\:\sqrt{\frac{\mathrm{1}\:+\:\mathrm{cos}\:{x}}{\mathrm{1}\:−\:\mathrm{cos}\:{x}}}\:=\:\mathrm{2}\:\centerdot\:\mathrm{cosec}\:{x} \\ $$$$\: \\ $$ Answered by Rasheed.Sindhi last updated on 14/Sep/22…
Question Number 45085 by Necxx last updated on 08/Oct/18 $${A}\:{motorist}\:{travelled}\:{from}\:{A}\:{to}\:{B}. \\ $$$${This}\:{is}\:{a}\:{distance}\:{of}\:\mathrm{142}{km}\:{at}\:{an} \\ $$$${average}\:{speed}\:{of}\:\mathrm{60}{kmhr}^{−\mathrm{1}} .{He} \\ $$$${spent}\:\mathrm{5}/\mathrm{2}{hours}\:{in}\:{B}\:{and}\:{then} \\ $$$${returned}\:{to}\:{A}\:{at}\:{an}\:{average}\:{speed} \\ $$$${of}\:\mathrm{80}{kmh}^{−\mathrm{1}} . \\ $$$$\left.{a}\right){At}\:{what}\:{time}\:{did}\:{the}\:{man}\:{arrive} \\…
Question Number 176140 by mnjuly1970 last updated on 13/Sep/22 $$ \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:,\:{prove}\:{that}: \\ $$$$\:\:{sin}\left(\mathrm{2}{A}\right)+{sin}\left(\mathrm{2}{B}\right)+{sin}\left(\mathrm{2}{C}\right)\: \\ $$$$\leqslant{cos}\:\left(\frac{{A}}{\mathrm{2}}\right)\:+\:{cos}\:\left(\frac{{B}}{\mathrm{2}}\right)+{cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$ \\ $$ Terms of Service Privacy…