Question Number 188128 by Rupesh123 last updated on 26/Feb/23 Answered by mr W last updated on 26/Feb/23 $${WLG},\:{assume}\:{a}\leqslant{b}\leqslant{c}. \\ $$$$\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{2}{b}×\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}}…
Question Number 57001 by nisha sinah last updated on 28/Mar/19 $${construct}\:{an}\:{analytic}\:{function}\:{f}\left({z}\right)\:{whose}\:{real}\:{part}\:{is}\:{e}^{{x}} \mathrm{cos}\:{y} \\ $$ Commented by 121194 last updated on 28/Mar/19 $${f}\left({z}\right)={e}^{{z}} \\ $$$${e}^{{z}} ={e}^{{x}+{iy}}…
Question Number 188048 by Rupesh123 last updated on 25/Feb/23 Answered by mr W last updated on 25/Feb/23 $${say}\:{A}={area},\:{r}={inradius} \\ $$$${A}=\frac{{ab}}{\mathrm{2}} \\ $$$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\…
Question Number 122505 by aupo14 last updated on 17/Nov/20 Answered by MJS_new last updated on 17/Nov/20 $${t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{5}\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}} \\ $$$${t}^{\mathrm{4}}…
Question Number 56954 by rahul 19 last updated on 27/Mar/19 $${Find}\:{minimum}\:{value}\:{of}\:: \\ $$$${cos}\left(\omega−\phi\right)+\mathrm{cos}\left(\phi−\varphi\right)+\mathrm{cos}\:\left(\varphi−\omega\right). \\ $$ Commented by rahul 19 last updated on 27/Mar/19 $${Ans}:−\frac{\mathrm{3}}{\mathrm{2}} \\…
Question Number 56681 by Joel578 last updated on 21/Mar/19 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{sin}\:\mid{x}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mathrm{tan}\:\mid{x}\mid\:\:\:\:\mathrm{for}\:\:\:\:\mid{x}\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by Abdo msup. last updated on 22/Mar/19 $${let}\:\mid{x}\mid={t}\:\:{let}\:{prove}\:{that}\:{sint}\leqslant{t}\leqslant{tant}\:{for}\mathrm{0}\leqslant{t}<\frac{\pi}{\mathrm{2}} \\ $$$${let}\:{W}\left({x}\right)={t}−{sint}\:\Rightarrow{W}^{,}…
Question Number 122118 by AbdullahMohammadNurusSafa last updated on 14/Nov/20 $$\frac{{tanA}}{\mathrm{1}−{cotA}}\:+\:\frac{{cotA}}{\mathrm{1}−{tanA}}\:=\:{secA}\:{cosecA}\:+\mathrm{1} \\ $$$$\boldsymbol{{Please}}\:\boldsymbol{{prove}}\:\boldsymbol{{it}}\:\boldsymbol{{for}}\:\boldsymbol{{me}}!!! \\ $$ Answered by MJS_new last updated on 14/Nov/20 $$\frac{{t}}{\mathrm{1}−\frac{\mathrm{1}}{{t}}}+\frac{\frac{\mathrm{1}}{{t}}}{\mathrm{1}−{t}}=\frac{{t}^{\mathrm{2}} }{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}\left({t}−\mathrm{1}\right)}={t}+\frac{\mathrm{1}}{{t}}+\mathrm{1}= \\ $$$$=\frac{{s}}{{c}}+\frac{{c}}{{s}}+\mathrm{1}=\frac{{s}^{\mathrm{2}}…
Question Number 122076 by peter frank last updated on 14/Nov/20 Answered by MJS_new last updated on 14/Nov/20 $$\mathrm{let}\:{t}=\mathrm{tan}\:{A} \\ $$$$\frac{\mathrm{1}+{t}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+{t}−\mathrm{1}}=\frac{\mathrm{1}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}−{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+{t}+\mathrm{1}} \\…
Question Number 122053 by 676597498 last updated on 13/Nov/20 $$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{E}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$$$\mathrm{E}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common} \\ $$$$\mathrm{interger}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by mr W last updated on 13/Nov/20 $${x}−\left[{x}\right]=\left\{{x}\right\}…
Question Number 187559 by Mingma last updated on 18/Feb/23 Answered by cortano12 last updated on 19/Feb/23 $$\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:.\frac{\mathrm{cos}\:{y}}{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{3tan}\:{x}=\mathrm{tan}\:{y} \\ $$$$\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{4tan}\:{x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\:=\pm\:\sqrt{\mathrm{15}} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}=\mathrm{sin}\:{y}\:;\:\mathrm{2cos}\:{x}=\mathrm{3cos}\:{y} \\…