Question Number 182848 by malithxd last updated on 15/Dec/22 $$\frac{{sin}^{\mathrm{3}} {x}−{sin}\:{x}+{cos}\:{x}}{{sin}\:{x}}={cot}\:{x}\:−{cos}^{\mathrm{2}} {x} \\ $$$${prove}\:{this} \\ $$$$ \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 182842 by malwan last updated on 15/Dec/22 $${prove}\:{that} \\ $$$${sec}\left({tan}^{−\mathrm{1}} {x}\right)=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$ Answered by cortano1 last updated on 15/Dec/22 $$\:{let}\:{x}\:=\:\mathrm{tan}\:{t}\: \\…
Question Number 117259 by bemath last updated on 10/Oct/20 Answered by Dwaipayan Shikari last updated on 10/Oct/20 $${tan}\alpha+{tan}\beta=\mathrm{3} \\ $$$${tan}\alpha{tan}\beta=−\mathrm{3} \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${sin}\left(\alpha+\beta\right)=\frac{\mathrm{3}}{\mathrm{5}}\:,{cos}\left(\alpha+\beta\right)=\frac{\mathrm{4}}{\mathrm{5}} \\…
Question Number 117257 by bemath last updated on 10/Oct/20 $${If}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)=\mathrm{1}\:{then}\: \\ $$$${what}\:{the}\:{value}\:{of}\:\mathrm{sin}\:^{\mathrm{11}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{11}} \left({x}\right)\:=? \\ $$ Commented by bemath last updated on 10/Oct/20…
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Question Number 117188 by bemath last updated on 10/Oct/20 $${Given}\:{r}+\frac{\mathrm{1}}{{r}}\:=\:\sqrt{\mathrm{2}}\:,\:{then}\:{r}^{\mathrm{8}} +\frac{\mathrm{1}}{{r}^{\mathrm{8}} }\:=\:?\: \\ $$ Commented by bemath last updated on 10/Oct/20 $${thank}\:{you}\:{sirs} \\ $$ Answered…
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Question Number 117144 by cantor last updated on 09/Oct/20 $$\boldsymbol{{S}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\prod}}\boldsymbol{{cos}}\left(\boldsymbol{{kx}}\right)=?????? \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}} \\ $$ Answered by Olaf last updated on 10/Oct/20 $$…
Question Number 117046 by 1549442205PVT last updated on 09/Oct/20 $$\mathrm{Calculate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{caculator}: \\ $$$$\left.\mathrm{a}\right)−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{sin10}°\left(\mathrm{2sin35}°−\frac{\mathrm{sec5}°}{\mathrm{2}}−\frac{\mathrm{cos40}°}{\mathrm{sin5}°}\right) \\ $$$$\left.\mathrm{b}\right)\mathrm{sin6}°−\mathrm{sin42}°−\mathrm{sin66}°+\mathrm{sin78}° \\ $$ Answered by bemath last updated on 09/Oct/20 $$\left(\mathrm{a}\right)\:−\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{35}°\:\mathrm{sin}\:\mathrm{10}°+\:\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{cos}\:\mathrm{5}°}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{40}°\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{5}°}\:= \\…
Question Number 51508 by 786786AM last updated on 27/Dec/18 $$\mathrm{A}\:\mathrm{man}\:\mathrm{walking}\:\mathrm{due}\:\mathrm{to}\:\mathrm{west}\:\mathrm{along}\:\mathrm{a}\:\mathrm{level}\:\mathrm{road}\:\mathrm{observes}\:\mathrm{a}\:\mathrm{school}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{N}\:\mathrm{72}°\:\mathrm{E}.\:\mathrm{After}\:\mathrm{walking}\:\mathrm{1500}\:\mathrm{yards},\: \\ $$$$\mathrm{he}\:\mathrm{observes}\:\mathrm{it}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{N}\:\mathrm{67}°\:\mathrm{E}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{is}\:\mathrm{the}\:\mathrm{school}\:\mathrm{from}\:\mathrm{the}\:\mathrm{road}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com