Question Number 58406 by rahul 19 last updated on 22/Apr/19 $$\left.\mathrm{1}\right){Value}\:{of}\:\mathrm{20}!+\frac{\mathrm{21}!}{\mathrm{1}!}+\frac{\mathrm{22}!}{\mathrm{2}!}+….+\frac{\mathrm{60}!}{\mathrm{40}!}\:{is}\:\:? \\ $$$$\left.\mathrm{2}\right)\:{Sum}\:{of}\:{all}\:{solutions}\:{of}\:{eq}^{{n}} \:: \\ $$$$\mathrm{cos}\:\mathrm{3}\theta=\mathrm{sin}\:\mathrm{2}\theta\:{in}\:{interval}\:\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\:{is}\:? \\ $$ Answered by MJS last updated on 23/Apr/19…
Question Number 189482 by MATHEMATICSAM last updated on 17/Mar/23 $$\mathrm{If}\:\mathrm{tan}\:\mathrm{11}°\:=\:{x}\:\mathrm{then}\:\mathrm{tan}\:\mathrm{1}°\:=\:?\: \\ $$ Answered by Frix last updated on 17/Mar/23 $$\mathrm{tan}\:\mathrm{11}°\:=\mathrm{tan}\:\left(\mathrm{9}°+\mathrm{2}°\right)\:={x} \\ $$$$=\frac{\mathrm{tan}\:\mathrm{9}°\:+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{9}°\:\mathrm{tan}\:\mathrm{2}°}={x} \\ $$$$\mathrm{tan}\:\mathrm{2}°\:=\frac{{x}−\mathrm{tan}\:\mathrm{9}°}{\mathrm{1}+{x}\mathrm{tan}\:\mathrm{9}°}={y} \\…
Question Number 189468 by cortano12 last updated on 17/Mar/23 $$\:\:\:\mathrm{If}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{csc}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{then} \\ $$$$\:\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=? \\ $$ Commented by mehdee42 last updated on 17/Mar/23 $${the}\:{given}\:{relation}\:{is}\:{wrong}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={cscx}−{cotx}…
Question Number 189463 by Rupesh123 last updated on 17/Mar/23 Answered by witcher3 last updated on 17/Mar/23 $$\mathrm{cos}\left(\mathrm{60}+\mathrm{20}\right) \\ $$$$\mathrm{cos}\left(\mathrm{40}\right)=\mathrm{cos}\left(\mathrm{60}−\mathrm{20}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{20}\right)+\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{20}\right)\right) \\ $$$$\mathrm{cos80}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{20}\right)−\sqrt{\mathrm{3}}\mathrm{sin}\left(\mathrm{20}\right)\right) \\ $$$$=\left(\left(\mathrm{3cos}\left(\mathrm{20}\right)−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 189428 by Rupesh123 last updated on 16/Mar/23 Answered by BaliramKumar last updated on 16/Mar/23 $${tanx}\:{tan}\left(\mathrm{60}°\:+\:{x}\right)\:{tan}\left(\mathrm{60}°\:−\:{x}\right)\: \\ $$$${tanx}\left[\frac{{tan}\mathrm{60}°\:+\:{tanx}}{\mathrm{1}−{tan}\mathrm{60}°\:{tanx}}\right]\left[\frac{{tan}\mathrm{60}°\:−\:{tanx}}{\mathrm{1}\:+\:{tan}\mathrm{60}°\:{tanx}}\right] \\ $$$${tanx}\left[\frac{\sqrt{\mathrm{3}}\:+\:{tanx}}{\mathrm{1}−\sqrt{\mathrm{3}}\:{tanx}}\right]\left[\frac{\sqrt{\mathrm{3}}\:−\:{tanx}}{\mathrm{1}\:+\:\sqrt{\mathrm{3}}\:{tanx}}\right] \\ $$$${tanx}\left[\frac{\left(\mathrm{3}\:−\:{tan}^{\mathrm{2}} {x}\right.}{\mathrm{1}−\mathrm{3}\:{tan}^{\mathrm{2}} {x}}\right]…
Question Number 189270 by Rupesh123 last updated on 14/Mar/23 Commented by maths_plus last updated on 14/Mar/23 $$\mathrm{good} \\ $$ Answered by HeferH last updated on…
Question Number 189256 by BaliramKumar last updated on 13/Mar/23 $${Prove}\:{that} \\ $$$$\mathrm{sin10}°\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−………..\infty}}}}}}}}}}}}} \\ $$ Answered by Frix last updated on 14/Mar/23 $$\mathrm{sin}\:\mathrm{10}°\:=\mathrm{sin}\:\frac{\pi}{\mathrm{18}} \\ $$$$\mathrm{First}\:\mathrm{let}'\mathrm{s}\:\mathrm{find}\:{x}=\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{x}}}} \\…
Question Number 189233 by a.lgnaoui last updated on 13/Mar/23 $$\mathrm{1}\bullet{Evaluer}\::\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right) \\ $$$$\mathrm{2}\bullet{En}\:{deduire}:\frac{\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right)}{\boldsymbol{{Aire}}\left(\boldsymbol{{ABCD}}\right)} \\ $$ Commented by a.lgnaoui last updated on 13/Mar/23 $$\left[\boldsymbol{{A}}\:'\boldsymbol{{B}}'\:\boldsymbol{{D}}\right]{alignes} \\ $$ Commented…
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Question Number 189125 by Rupesh123 last updated on 12/Mar/23 Answered by cortano12 last updated on 12/Mar/23 $$\:\mathrm{cos}\:\mathrm{18}°=\mathrm{2cos}^{\mathrm{2}} \:\mathrm{9}°−\mathrm{1} \\ $$$$\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{9}°=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{18}°}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{18}°=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{18}°} \\…