Question Number 174556 by mnjuly1970 last updated on 04/Aug/22 Commented by infinityaction last updated on 04/Aug/22 $$\:\:\:\:\:\:\:\mathrm{tan}\alpha\:+\mathrm{3}\:=\:\mathrm{2sec}\alpha \\ $$$$\:\:\:\:\:\:\mathrm{3tan}^{\mathrm{2}} \alpha\:−\mathrm{6tan}\alpha\:−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{tan}\alpha\:\:=\:\:\:\frac{\mathrm{6}\:\pm\:\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{6}}\:\:\:=\:\:\frac{\mathrm{sin}\alpha\:}{\mathrm{cos}\alpha\:}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{sin}\alpha−\mathrm{cos}\alpha\:\:}{\mathrm{sin}\alpha+\mathrm{cos}\alpha\:\:}\:\:=\:\:\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{12}+\mathrm{4}\sqrt{\mathrm{6}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \\…
Question Number 43467 by peter frank last updated on 11/Sep/18 $${prove}\:{th}\mathrm{at}\:\mathrm{tan}\theta+\mathrm{2}{tan}\mathrm{2}\theta+\mathrm{4}{tan}\:\:\:\mathrm{4}\theta \\ $$$$+\mathrm{8cot8}\theta={cot}\theta \\ $$ Answered by ajfour last updated on 11/Sep/18 $$\mathrm{8cot}\:\mathrm{8}\theta+\mathrm{4tan}\:\mathrm{4}\theta+\mathrm{2tan}\:\mathrm{2}\theta+\mathrm{tan}\:\theta \\ $$$$\:\:\:\:=\:\frac{\mathrm{8}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}}…
Question Number 174435 by cortano1 last updated on 01/Aug/22 $$\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{3}°+…+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}°=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 43346 by peter frank last updated on 10/Sep/18 $$\mathrm{sin}\:{x}−\mathrm{sin}\:\mathrm{5}{x}=\mathrm{sin}\:\mathrm{3}{x}\:{find}\:{the}\:{angle} \\ $$$${that}\:{satisfied}\:{the}\:{equestion} \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 $${sinx}={sin}\mathrm{5}{x}+{sin}\mathrm{3}{x}…
Question Number 174346 by cortano1 last updated on 30/Jul/22 $$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°}\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}°}\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}°}\:−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{45}°\:}=? \\ $$ Answered by behi834171 last updated on 30/Jul/22 $$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{10}}=\mathrm{1}+{tg}^{\mathrm{2}}…
Question Number 43268 by Rauny last updated on 09/Sep/18 $$\mathrm{probably},\:\mathrm{cos}\:{nx}=\mathrm{2}^{{n}−\mathrm{1}} \mathrm{cos}^{{n}} \:{x}−{n}\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}^{{n}−\mathrm{2}} \: \\ $$$$+\frac{\left({n}−\mathrm{3}\right){n}}{\mathrm{2}}\mathrm{2}^{{n}−\mathrm{5}} \mathrm{cos}^{{n}−\mathrm{4}} \:{x}\ldots \\ $$$$\mathrm{wow} \\ $$ Commented by Rauny…
Question Number 43267 by Rauny last updated on 09/Sep/18 $$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}{x}=\mathrm{4cos}^{\mathrm{3}} \:{x}−\mathrm{3cos}\:{x} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}=\mathrm{8cos}^{\mathrm{4}} \:{x}−\mathrm{8cos}^{\mathrm{2}} \:{x}+\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{5}{x}=\mathrm{16cos}^{\mathrm{5}} \:{x}−\mathrm{20cos}^{\mathrm{3}} +\mathrm{5cos}\:{x}\: \\ $$$$\mathrm{cos}\:\mathrm{6}{x}=\mathrm{32cos}^{\mathrm{6}} \:{x}−\mathrm{48cos}^{\mathrm{4}}…
Question Number 43266 by gunawan last updated on 09/Sep/18 $$\mathrm{cos}^{\mathrm{3}} {x}+\mathrm{cos}^{−\mathrm{3}} {x}=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}=… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18 $${t}={cosx}+{isinx}={e}^{{ix}} \\…
Question Number 108780 by 150505R last updated on 19/Aug/20 Answered by bobhans last updated on 19/Aug/20 $$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{81}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{9}\: \\ $$$$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{63}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{27}\: \\ $$$$\left(\ast\right)\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cot}\:^{\mathrm{2}}…
Question Number 43192 by Raj Singh last updated on 08/Sep/18 Commented by Rauny last updated on 09/Sep/18 $$\mathrm{cosec}\:{x}+\mathrm{sec}\:{x}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\mathrm{2} \\ $$$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}=\mathrm{2sin}\:{x}\mathrm{cos}\:{x} \\ $$$${A}={B}\Rightarrow{A}^{\mathrm{2}} ={B}^{\mathrm{2}} , \\…