Question Number 169709 by alcohol last updated on 06/May/22 Commented by alcohol last updated on 06/May/22 please help me solve anyone you can please thanks so much Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 169702 by Anyangwa last updated on 06/May/22 $$\begin{vmatrix}{\mathrm{0}}&{\mathrm{4}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{\mathrm{5}}&{\mathrm{1}}\end{vmatrix}=\: \\ $$ Answered by MikeH last updated on 06/May/22 $$\mathrm{Identify}\:\mathrm{the}\:\mathrm{row}\:\mathrm{with}\:\mathrm{maximum}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{zeros}. \\ $$$$−\mathrm{4}\left(\mathrm{4}\left(\mathrm{2}−\mathrm{5}\right)+\left(\mathrm{15}−\mathrm{4}\right)\right)+\left(\mathrm{4}\left(\mathrm{5}−\mathrm{2}\right)−\mathrm{0}+\left(\mathrm{6}−\mathrm{10}\right)\right. \\…
Question Number 38618 by tawa tawa last updated on 27/Jun/18 Commented by MrW3 last updated on 28/Jun/18 $$\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{x}\right)\right]^{\mathrm{4}} =\mathrm{1}−\mathrm{4}{x}+{px}^{\mathrm{2}} +{qx}^{\mathrm{3}} +… \\ $$$${this}\:{can}\:{not}\:{be}\:{true}. \\ $$$$\Rightarrow{something}\:{is}\:{wrong}\:{with}\:{the}\:{question}.…
Question Number 38600 by ajfour last updated on 27/Jun/18 Commented by ajfour last updated on 27/Jun/18 $${Find}\:\theta\:{in}\:{terms}\:{of}\:\boldsymbol{{c}}\:. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 104122 by Dwaipayan Shikari last updated on 19/Jul/20 $$\frac{\mathrm{1}}{\mathrm{sin2x}}+\frac{\mathrm{1}}{\mathrm{sin2}^{\mathrm{2}} \mathrm{x}}+…..+\frac{\mathrm{1}}{\mathrm{sin2}^{\mathrm{n}} \mathrm{x}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\: \\ $$ Answered by OlafThorendsen last updated on 19/Jul/20 $$\frac{\mathrm{1}}{\mathrm{sin}{u}}\:=\:\frac{\mathrm{sin}\left({u}−\frac{{u}}{\mathrm{2}}\right)}{{sinu}\mathrm{sin}\frac{{u}}{\mathrm{2}}}…
Question Number 104016 by Dwaipayan Shikari last updated on 18/Jul/20 Answered by OlafThorendsen last updated on 18/Jul/20 $$\mathrm{arctan}{x}−\mathrm{arctan}{y}\:=\:\mathrm{arctan}\frac{{x}−{y}}{\mathrm{1}+{xy}} \\ $$$$\mathrm{arctan}\left(\mathrm{2}{k}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{2}{k}−\mathrm{1}\right)\:=\:\mathrm{arctan}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)−\left(\mathrm{2}{k}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$=\:\:\mathrm{arctan}\frac{\mathrm{2}}{\mathrm{4}{k}^{\mathrm{2}} }\:=\:\mathrm{arctan}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} } \\…
Question Number 169537 by cortano1 last updated on 02/May/22 Commented by infinityaction last updated on 02/May/22 $${what}\:{is}\:{halle}\:?? \\ $$ Commented by cortano1 last updated on…
Question Number 103984 by dw last updated on 18/Jul/20 $$\:\:\:\:\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{tan}}^{\mathrm{2}} \left(\mathrm{2}\boldsymbol{{x}}\right)+\boldsymbol{{tan}}^{\mathrm{2}} \left(\mathrm{4}\boldsymbol{{x}}\right)=\mathrm{33} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=? \\ $$ Commented by behi83417@gmail.com last updated on 19/Jul/20 $$\mathrm{Nice}\:\mathrm{to}\:\mathrm{see}\:\mathrm{you}\:\mathrm{again}\:\mathrm{mr}\:\mathrm{proph}:\mathrm{MJS}.…
Question Number 103887 by bemath last updated on 18/Jul/20 $${how}\:{do}\:{you}\:{prove}\:\mathrm{sin}\:\left({a}+{b}\right)\:=\:\mathrm{sin} \\ $$$${a}\:\mathrm{cos}\:{b}\:+\:\mathrm{cos}\:{a}\:\mathrm{sin}\:{b} \\ $$$${geometrically}\:? \\ $$ Commented by $@y@m last updated on 18/Jul/20 https://medium.com/@nubtrek/easy-with-sin-a-b-proof-vuja-de-50c15fca3344 Commented…
Question Number 38288 by ajfour last updated on 23/Jun/18 $$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\:=\:\mathrm{2tan}\:\theta \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$${express}\:\mathrm{sin}\:\alpha,\:\mathrm{sin}\:\beta\:\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18 $${t}_{\mathrm{1}} ={tan}\frac{\alpha}{\mathrm{2}}\:\:\:{t}_{\mathrm{2}}…