Question Number 54755 by Knight last updated on 10/Feb/19 $${solve}\:{for}\:{x} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:=\:\mathrm{1}\mp\sqrt{\mathrm{2}\:}\:,\:\mp\mathrm{1}\:,−\mathrm{1}\mp\sqrt{\mathrm{2}}\:\:\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 185821 by Rupesh123 last updated on 28/Jan/23 Answered by qaz last updated on 28/Jan/23 $$\mathrm{5arctan}\:\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{2arctan}\:\frac{\mathrm{3}}{\mathrm{79}}=\mathrm{5}{Arg}\left(\mathrm{7}+{i}\right)+\mathrm{2}{Arg}\left(\mathrm{79}+\mathrm{3}{i}\right) \\ $$$$={Arg}\left[\left(\mathrm{7}+{i}\right)^{\mathrm{5}} \centerdot\left(\mathrm{79}+\mathrm{3}{i}\right)^{\mathrm{2}} \right]={Arg}\left[\mathrm{78125000}\left(\mathrm{1}+{i}\right)\right] \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{78125000}}{\mathrm{78125000}}=\frac{\pi}{\mathrm{4}} \\ $$…
Question Number 185794 by Rupesh123 last updated on 27/Jan/23 Commented by Rupesh123 last updated on 27/Jan/23 Solve in R Answered by a.lgnaoui last updated on 27/Jan/23 $$\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=−\mathrm{1}\:\:\:\:\:{y}=−{x}+{k}\pi\:\:\:\:\left({k}\in\mathbb{Z}\right)…
Question Number 54721 by 951172235v last updated on 09/Feb/19 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\Sigma\left\{\mathrm{sin}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}}{\Sigma\left\{\mathrm{cos}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}\:\:\:}\:\:=\:\mathrm{tan}\:\mathrm{2xtan}\:\mathrm{2ytan}\:\mathrm{2z} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 120204 by bemath last updated on 30/Oct/20 $${Call}\:{a}\:\mathrm{7}−{digit}\:{telephone}\:{number}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} −{d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \\ $$$${memorable}\:{if}\:{the}\:{prefix}\:{sequence}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} \\ $$$${is}\:{exactly}\:{the}\:{same}\:{as}\:{either}\:{of}\:{the}\:{sequences} \\ $$$${d}_{\mathrm{4}} {d}_{\mathrm{5}}…
Question Number 185732 by Rupesh123 last updated on 26/Jan/23 Answered by witcher3 last updated on 26/Jan/23 $$\mathrm{tg}\left(\mathrm{a}\right)−\mathrm{tg}\left(\mathrm{b}\right)=\frac{\mathrm{sin}\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right)} \\ $$$$\mathrm{tg}\left(\mathrm{50}\right)−\mathrm{tg}\left(\mathrm{40}\right)=\frac{\mathrm{sin}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)}=\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)}=\mathrm{2}\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{2cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)} \\ $$$$=\frac{\mathrm{2cos}\left(\mathrm{10}\right)}{\mathrm{sin}\left(\mathrm{80}\right)}=\mathrm{2} \\ $$…
Question Number 185731 by Mingma last updated on 26/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54626 by Aditya789 last updated on 08/Feb/19 $$\frac{\mathrm{1}+\mathrm{sinx}+\mathrm{cox}}{\mathrm{1}+\mathrm{sinx}−\mathrm{cosx}}=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{1}+\mathrm{cosx}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19 $${LHS} \\ $$$$\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}} \\…
Question Number 120156 by mnjuly1970 last updated on 29/Oct/20 $$\:\:\:\:\:\:\:\:…\spadesuit\:{nice}\:\:{calculus}\spadesuit… \\ $$$$\:\:{if}\:\:{cos}^{−\mathrm{1}} \left({x}\right)+{cos}^{−\mathrm{1}} \left({y}\right)+{cos}^{−\mathrm{1}} \left({z}\right)=\pi\checkmark \\ $$$$ \\ $$$$\:\:\:\:\:\:{show}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xyz}=\mathrm{1}\:\:\checkmark \\…
Question Number 185647 by Mingma last updated on 24/Jan/23 Answered by mahdipoor last updated on 24/Jan/23 $${secx}+{tan}^{\mathrm{3}} {x}.{cscx}=\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}=\frac{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}= \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{3}}…