Question Number 148773 by mim24 last updated on 31/Jul/21 Answered by som(math1967) last updated on 31/Jul/21 $$\frac{\mathrm{1}}{\boldsymbol{{sin}}\mathrm{20}}\:+\frac{\sqrt{\mathrm{3}}}{\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\boldsymbol{{cos}}\mathrm{20}+\sqrt{\mathrm{3}}\boldsymbol{{sin}}\mathrm{20}}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}\mathrm{20}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{sin}}\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}.\mathrm{2}\left(\boldsymbol{{cos}}\mathrm{60}\boldsymbol{{cos}}\mathrm{20}+\boldsymbol{{sin}}\mathrm{60}\boldsymbol{{sin}}\mathrm{20}\right)}{\mathrm{2}\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\left(\mathrm{60}−\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{40}}…
Question Number 148775 by Jonathanwaweh last updated on 31/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148768 by mim24 last updated on 31/Jul/21 Answered by liberty last updated on 31/Jul/21 $$\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{120}°+\mathrm{2A}\right)+\mathrm{1}−\mathrm{cos}\:\mathrm{2A}+\mathrm{1}−\mathrm{cos}\:\left(\mathrm{120}°−\mathrm{2A}\right)}{\mathrm{2}}\:= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{\mathrm{cos}\:\left(\mathrm{120}°+\mathrm{2A}\right)+\mathrm{cos}\:\left(\mathrm{120}°−\mathrm{2A}\right)\right\}}{\mathrm{2}}= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{\mathrm{2cos}\:\mathrm{120}°\mathrm{cos}\:\mathrm{2A}\right\}}{\mathrm{2}}= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{−\mathrm{cos}\:\mathrm{2A}\right\}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$…
Question Number 83235 by peter frank last updated on 28/Feb/20 Commented by peter frank last updated on 28/Feb/20 $${c}\:{help} \\ $$ Commented by john santu…
Question Number 148767 by mim24 last updated on 31/Jul/21 Answered by liberty last updated on 31/Jul/21 $$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{20}°}=\frac{\mathrm{cos}\:\mathrm{20}°+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{20}°\mathrm{cos}\:\mathrm{20}°} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}°+\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{20}°}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{40}°} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{30}°\mathrm{cos}\:\mathrm{20}°+\mathrm{cos}\:\mathrm{30}°\mathrm{sin}\:\mathrm{20}°}{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{40}°} \\ $$$$=\frac{\mathrm{4sin}\:\mathrm{50}°}{\mathrm{cos}\:\mathrm{50}°}=\mathrm{4tan}\:\mathrm{50}°\:=\:\mathrm{4cot}\:\mathrm{40}° \\ $$…
Question Number 83229 by peter frank last updated on 28/Feb/20 Commented by jagoll last updated on 28/Feb/20 $$\left(\mathrm{iii}\right)\:\frac{\mathrm{cos}\:\mathrm{3A}−\mathrm{cos}\:\mathrm{9A}+\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{3A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{3A}+\mathrm{sin}\:\mathrm{3A}−\mathrm{sin}\:\mathrm{A}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{9A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{A}}\:=\:\frac{\mathrm{2sin}\:\mathrm{5Asin}\:\mathrm{4A}}{\mathrm{2cos}\:\mathrm{5Asin}\:\mathrm{4A}} \\ $$$$=\:\mathrm{tan}\:\mathrm{5A} \\ $$ Commented…
Question Number 17692 by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17 Answered by Tinkutara last updated on 09/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17 $${thanks}\:{sir}.{nice}\:{and}\:{smart}. \\…
Question Number 17689 by Tinkutara last updated on 09/Jul/17 $$\mathrm{A}\:\mathrm{bird}\:\mathrm{is}\:\mathrm{tossing}\:\left(\mathrm{flying}\:\mathrm{to}\:\mathrm{and}\:\mathrm{fro}\right) \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{cars}\:\mathrm{moving}\:\mathrm{towards} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}.\:\mathrm{One}\:\mathrm{car} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{18}\:\mathrm{km}/\mathrm{h}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{27}\:\mathrm{km}/\mathrm{h}.\:\mathrm{The}\:\mathrm{bird} \\ $$$$\mathrm{starts}\:\mathrm{moving}\:\mathrm{from}\:\mathrm{first}\:\mathrm{car}\:\mathrm{towards} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{and}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{the}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{36}\:\mathrm{km}/\mathrm{h}\:\mathrm{and}\:\mathrm{when}\:\mathrm{the}\:\mathrm{two}\:\mathrm{cars}\:\mathrm{were} \\…
Question Number 148756 by mathdanisur last updated on 30/Jul/21 $$\begin{cases}{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{x}\:+\:{y}\:=\:−\mathrm{2}}\\{{y}^{\mathrm{2}} \:+\:{xy}\:+\:{x}\:+\:{y}\:=\:\mathrm{1}}\end{cases}\:\:\Rightarrow\:\mathrm{3}{x}\:+\:{y}\:=\:? \\ $$ Answered by mindispower last updated on 30/Jul/21 $$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+\left({x}+{y}\right)\right)^{\mathrm{2}}…
Question Number 148753 by Jonathanwaweh last updated on 30/Jul/21 Answered by mathmax by abdo last updated on 31/Jul/21 $$\mathrm{A}_{\mathrm{n}} =\prod_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \:\mathrm{e}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)^{\mathrm{k}^{\mathrm{2}} } \:\Rightarrow\mathrm{A}_{\mathrm{n}}…