Question Number 148723 by Rustambek last updated on 30/Jul/21 $${tgx}+{tgy}=\mathrm{4}\:\:\:{cosx}+{cosy}=\frac{\mathrm{1}}{\mathrm{5}}\:\:\: \\ $$$${tg}\left({x}+{y}\right)=? \\ $$ Commented by Rustambek last updated on 30/Jul/21 $${please} \\ $$ Terms…
Question Number 17647 by Tinkutara last updated on 09/Jul/17 $${ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangular}\:\mathrm{park}\:\mathrm{with}\:{AB}\:= \\ $$$${AC}\:=\:\mathrm{100}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{clock}\:\mathrm{tower}\:\mathrm{is}\:\mathrm{situated} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{BC}.\:\mathrm{The}\:\mathrm{angles}\:\mathrm{of} \\ $$$$\mathrm{elevation}\:\mathrm{of}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}\:\mathrm{at}\:{A}\:\mathrm{and} \\ $$$${B}\:\mathrm{are}\:\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{3}.\mathrm{2}\right)\:\mathrm{and}\:\mathrm{cosec}^{−\mathrm{1}} \left(\mathrm{2}.\mathrm{6}\right) \\ $$$$\mathrm{respectively}.\:\mathrm{The}\:\mathrm{height}\:\mathrm{of}\:\mathrm{tower}\:\mathrm{is} \\ $$ Commented…
Question Number 17645 by Tinkutara last updated on 09/Jul/17 $$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:{M}\:\mathrm{lying}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallelogram}\:{ABCD}, \\ $$$$\mathrm{two}\:\mathrm{parallels}\:\mathrm{to}\:{AB}\:\mathrm{and}\:{AD}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:{ABCD}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{P},\:{Q},\:{R},\:{S}\:\left(\mathrm{See}\:\mathrm{Figure}\right).\:\mathrm{Prove} \\ $$$$\mathrm{that}\:{M}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{diagonal}\:{AC}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\left[{MRDS}\right]\:=\:\left[{MPBQ}\right]. \\ $$ Commented…
Question Number 148708 by mathdanisur last updated on 30/Jul/21 $${tg}^{\mathrm{6}} \left(\mathrm{36}°\right)\:+\:{tg}^{\mathrm{6}} \left(\mathrm{72}°\right)\:=\:? \\ $$ Commented by liberty last updated on 31/Jul/21 $$\mathrm{850} \\ $$ Terms…
Question Number 17638 by chux last updated on 08/Jul/17 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{this} \\ $$$$\mathrm{confusing}\:\mathrm{question} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2x}/\mathrm{y}} ×\mathrm{y}^{\mathrm{y}/\mathrm{x}} =\mathrm{4}……\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\left(\mathrm{xy}\right)^{\mathrm{xy}+\mathrm{yx}} =\mathrm{16}…..\left(\mathrm{2}\right) \\ $$$$…
Question Number 148711 by Rankut last updated on 30/Jul/21 $${if}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{4},\:{and}\:{x}−\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$${then}\:{prove}\:{that}\:\mathrm{4}=\mathrm{5} \\ $$ Answered by liberty last updated on 30/Jul/21 $$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{16}\Rightarrow\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 17634 by tawa tawa last updated on 08/Jul/17 $$\mathrm{y}\:=\:\mathrm{x}!\:\:,\:\:\:\:\:\mathrm{Find}\:\:\:\mathrm{y}' \\ $$ Answered by alex041103 last updated on 09/Jul/17 $${Because}\:{x}!\:{is}\:{defined}\:{only}\:{for}\:{non}−{negative} \\ $$$${integers},\:{i}.{e}.\:{the}\:{function}\:{isn}'{t}\:{continious} \\ $$$${we}\:{cannot}\:{intagrate}\:{or}\:{diferentiate}\:{it}.…
Question Number 148707 by mathdanisur last updated on 30/Jul/21 $${Solve}\:{for}\:{equation}: \\ $$$$\mathrm{2}{tg}\left(\mathrm{3}{x}\right)\:-\:\mathrm{3}{tg}\left(\mathrm{2}{x}\right)\:=\:{tg}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:\centerdot\:{tg}\left(\mathrm{3}{x}\right) \\ $$ Answered by nimnim last updated on 30/Jul/21 $${Let}\:{me}\:{give}\:{a}\:{try}…. \\ $$$$\Rightarrow\mathrm{2}\left(\frac{\mathrm{3}{tanx}−{tan}^{\mathrm{3}}…
Question Number 148706 by mathdanisur last updated on 30/Jul/21 Answered by mindispower last updated on 30/Jul/21 $${sin}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left({E}\right)\Leftrightarrow\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)+{cos}\left(\mathrm{14}{x}\right)\right)=\mathrm{2} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{14}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{4}{x}\right)+\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{2}{x}\right)=\mathrm{0} \\…
Question Number 83166 by john santu last updated on 28/Feb/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{x}^{\mathrm{5}\:} \mathrm{in}\:\mathrm{expansion} \\ $$$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} ×\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} \: \\ $$ Commented by mr W last updated…