Question Number 191615 by MATHEMATICSAM last updated on 27/Apr/23 $${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that}, \\ $$$$\frac{{a}}{{a}^{\mathrm{2}} \:−\:{bc}}\:+\:\frac{{b}}{{b}^{\mathrm{2}} \:−\:{ca}}\:+\:\frac{{c}}{{c}^{\mathrm{2}} \:−\:{ab}}\:=\:\mathrm{0}. \\ $$ Answered by som(math1967) last updated on 27/Apr/23 $$\:{a}+{b}+{c}=\mathrm{0}…
Question Number 191614 by vishal1234 last updated on 27/Apr/23 $$\frac{\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} }{\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} }\:=\: \\ $$$$\frac{\left(−{w}\right)^{\mathrm{100}} +\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} }{\left(−{w}\right)^{\mathrm{100}} −\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} } \\ $$$$=\:\frac{{w}^{\mathrm{100}} +{w}^{\mathrm{200}}…
Question Number 191610 by mehdee42 last updated on 27/Apr/23 $${Q}:\:{if}\:\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}{cos}\left(\theta\right)\:\:{prove}\:{it}\:\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\mathrm{2}{cos}\left({n}\theta\right) \\ $$ Answered by Frix last updated on 27/Apr/23 $${z}+\frac{\mathrm{1}}{{z}}=\mathrm{2}{c}\:\Rightarrow\:{z}={c}\pm\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${x}=\mathrm{cos}\:\theta\:\pm\mathrm{i}\:\mathrm{sin}\:\theta\:=\mathrm{e}^{\pm\mathrm{i}\theta}…
Question Number 60533 by Tawa1 last updated on 21/May/19 $$\mathrm{If}\:\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{C}\:+\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}\:−\:\mathrm{B}\right)\:\:=\:\:\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{A}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B} \\ $$ Commented by malwaan last updated on 21/May/19 $$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}^{°} \\ $$$${cos}\left({x}+{y}\right)−{cos}\left({x}−{y}\right)=−\mathrm{2}{sin}\left({x}\right){sin}\left({y}\right) \\…
Question Number 191595 by MATHEMATICSAM last updated on 26/Apr/23 $${a}\:=\:\frac{{xy}}{{x}\:+\:{y}}\:,\:{b}\:=\:\frac{{xz}}{{x}\:+\:{z}}\:\mathrm{and}\:{c}\:=\:\frac{{yz}}{{y}\:+\:{z}}\:. \\ $$$$\mathrm{Represent}\:{x}\:\mathrm{in}\:{a},\:{b},\:{c}\:\mathrm{form}.\:\left[{x},\:{y},\:{z}\:\neq\:\mathrm{0}\right] \\ $$ Answered by mehdee42 last updated on 26/Apr/23 $$\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:\:\:\left({i}\right)\:,\:\:\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}\:\:\left({ii}\right)\:\:,\:\:\:\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\:\:\left({iii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\overset{\left({iii}\right)} {\Rightarrow}\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{c}}…
Question Number 191589 by MATHEMATICSAM last updated on 28/Apr/23 $${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$ Commented by mr W…
Question Number 60500 by prof Abdo imad last updated on 21/May/19 $${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{e}^{{A}} \:\:\:{and}\:{e}^{−\mathrm{2}{A}} \:. \\ $$ Commented by maxmathsup…
Question Number 60501 by prof Abdo imad last updated on 21/May/19 $${let}\:{A}\:=\begin{pmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right){calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{e}^{{A}} \:\:\:{and}\:{e}^{−{A}} \:. \\ $$$$ \\ $$ Commented by maxmathsup…
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Question Number 126030 by Fareed last updated on 16/Dec/20 $$\mathrm{2}^{\mathrm{x}} =\mathrm{4x} \\ $$$$\mathrm{solve}\:\mathrm{it}\:\mathrm{please} \\ $$ Answered by Dwaipayan Shikari last updated on 16/Dec/20 $$\mathrm{2}^{{x}} =\mathrm{4}{x}…