Question Number 167502 by cortano1 last updated on 18/Mar/22 $$\:\:\:\:\:\:\begin{cases}{\mathrm{tan}\:\alpha=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}}\\{\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}\end{cases}\Rightarrow\alpha+\beta=? \\ $$ Answered by Rasheed.Sindhi last updated on 18/Mar/22 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\mathrm{tan}\:\beta}\: \\ $$$$=\frac{\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}{\mathrm{1}−\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}=\frac{\mathrm{m}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{m}+\mathrm{1}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{2m}+\mathrm{1}\right)−\mathrm{m}} \\ $$$$=\frac{\mathrm{2m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{1}}{\mathrm{2m}^{\mathrm{2}}…
Question Number 167486 by greogoury55 last updated on 18/Mar/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{cos}\:\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi \\ $$ Answered by mr W last updated on 18/Mar/22 $$\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{6}{x}}{\mathrm{3}}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\…
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Question Number 167443 by infinityaction last updated on 16/Mar/22 Answered by nimnim last updated on 16/Mar/22 $$\:\:\:\:\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\right)+\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \mathrm{cos}\theta=\mathrm{1}…
Question Number 36338 by tanmay.chaudhury50@gmail.com last updated on 31/May/18 $${prove}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)…{upto}\:{infinity} \\ $$$$=\frac{\mathrm{1}}{\Pi}{sinh}\Pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 101713 by I want to learn more last updated on 04/Jul/20 $$\mathrm{Let}\:\:\:\mathrm{a}\:\:\mathrm{and}\:\:\mathrm{b}\:\:\mathrm{be}\:\mathrm{positive}\:\mathrm{numbers}\:\mathrm{satisfying}\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} \:\:=\:\:\mathrm{5}, \\ $$$$\mathrm{If}\:\:\:\:\mathrm{a}\:\mathrm{cos}\left(\theta\right)\:\:−\:\:\mathrm{b}\:\mathrm{sin}\left(\theta\right)\:\:=\:\:\mathrm{1},\:\:\:\:\:\mathrm{find}\:\:\:\:\mathrm{a}\:\mathrm{sin}\left(\theta\right)\:\:+\:\:\mathrm{b}\left(\mathrm{cos}\theta\right) \\ $$ Answered by mr W last…
Question Number 101590 by student work last updated on 03/Jul/20 $$\mathrm{p}\left(\mathrm{2x}+\mathrm{5}\right)=\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{1}\right)\mathrm{Q}\left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{if}\:\mathrm{Q}\left(−\mathrm{1}\right)=\mathrm{3}\:\:\:\:\:\mathrm{then}\:\mathrm{p}\left(\mathrm{1}\right)=? \\ $$ Commented by Tinku Tara last updated on 03/Jul/20 $$\mathrm{You}\:\mathrm{couldnt}\:\mathrm{simple}\:\mathrm{have}\:\mathrm{doubt}…
Question Number 101589 by student work last updated on 03/Jul/20 $$\mathrm{if}\:\mathrm{sin10}^{\mathrm{0}} =\mathrm{x}\:\:\:\:\:\:\:\:\:\:\mathrm{then}\:\:\:\mathrm{sin70}^{\mathrm{0}} =? \\ $$ Answered by Dwaipayan Shikari last updated on 03/Jul/20 $${sin}\mathrm{70}°={cos}\mathrm{20}°=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{10}°=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}}…
Question Number 101554 by student work last updated on 03/Jul/20 $$\frac{\mathrm{1}}{\mathrm{cos80}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{sin80}}=? \\ $$ Answered by bramlex last updated on 03/Jul/20 $$\frac{\mathrm{sin}\:\mathrm{80}^{\mathrm{o}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{80}^{\mathrm{o}} }{\mathrm{sin}\:\mathrm{80}^{\mathrm{o}} \mathrm{cos}\:\mathrm{80}^{\mathrm{o}} }\:=\:…
Question Number 101522 by bramlex last updated on 03/Jul/20 $$\mathrm{calculate}\::\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{47}^{\mathrm{o}} +\mathrm{cos}\:^{\mathrm{2}} \mathrm{73}^{\mathrm{o}} +\mathrm{cos}\:\mathrm{47}^{\mathrm{o}} .\mathrm{cos}\:\mathrm{73}^{\mathrm{o}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by bemath last updated on 03/Jul/20…