Question Number 25763 by amankumar last updated on 14/Dec/17 $$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right) \\ $$ Answered by ajfour last updated on 15/Dec/17 $$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}}…
Question Number 156807 by amin96 last updated on 15/Oct/21 $${f}\left({x}\right)={arctg}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\:{and}\:\alpha={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+\ldots+{f}\left(\mathrm{21}\right) \\ $$$${find}\:\:{tg}\left(\alpha\right)=? \\ $$ Commented by amin96 last updated on 15/Oct/21 $$\left.{A}\left.\right)\left.\frac{\mathrm{6}}{\mathrm{11}}\left.\:\left.\:\:\:{B}\right)\frac{\mathrm{7}}{\mathrm{11}}\:\:\:{C}\right)\frac{\mathrm{11}}{\mathrm{21}}\:\:\:\:{D}\right)\frac{\mathrm{1}}{\mathrm{21}}\:\:\:\:{E}\right)\frac{\mathrm{21}}{\mathrm{23}} \\ $$…
Question Number 25728 by ajfour last updated on 13/Dec/17 Commented by ajfour last updated on 13/Dec/17 $${If}\:{ratio}\:{of}\:{areas}\:{of}\:\bigtriangleup{PQR}\:{and} \\ $$$$\bigtriangleup{ABC}\:{be}\:{independent}\:{of}\:{a},{b},{c}\:; \\ $$$${find}\:{the}\:{ratio}. \\ $$ Commented by…
Question Number 25611 by nnnavendu last updated on 12/Dec/17 $${if}\:{cosh}\left({u}+{iv}\right)={x}+{iy}\:,\:{show}\:{that} \\ $$$$\frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} {u}}+\frac{{y}^{\mathrm{2}} }{{sinh}^{\mathrm{2}} {u}}=\mathrm{1}\:\:{and}\:\:\frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} {v}}−\frac{{y}^{\mathrm{2}} }{{sinh}^{\mathrm{2}} {v}}=\mathrm{1} \\ $$$$ \\ $$$${where}\:\:\:\:{sinhx}\:{and}\:{coshx}\:\:\:\:{hhyperbolcfunction} \\…
Question Number 90876 by john santu last updated on 26/Apr/20 $$\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{9}}\right)}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)}\:=\:? \\ $$ Commented by john santu last updated on 26/Apr/20 $${Ramanujan}\:{theorem} \\ $$$${let}\:\alpha,\:\beta\:,\gamma\:{be}\:{a}\:{roots}\: \\…
Question Number 25343 by Tinkutara last updated on 08/Dec/17 Commented by sushmitak last updated on 08/Dec/17 $${As}\:{pointed}\:{out}\:{below}.\:{This} \\ $$$${is}\:{incorrect}\:{logic}. \\ $$$${The}\:{correct}\:{logic}\:{for}\:{interchangable} \\ $$$${variables}\:{in}\:{an}\:{equation}\:{is}\:{that} \\ $$$${if}\:{p}\:{is}\:{a}\:{solutoon}\:{for}\:{b}\:{then}\:{it}…
Question Number 156321 by cortano last updated on 10/Oct/21 $$\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)=? \\ $$ Commented by john_santu last updated on 10/Oct/21 $${ans}\::\:\frac{\mathrm{19}}{\mathrm{16}} \\…
Question Number 156308 by cortano last updated on 10/Oct/21 $$\mathrm{Find}\:\mathrm{exact}\:\mathrm{form}\:\mathrm{sin}\:\mathrm{1}° \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 90739 by ajfour last updated on 25/Apr/20 $${solve}\:{for}\:{x}\:{and}\:{y} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]+\mathrm{cot}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$ Commented by me2love2math last updated on…
Question Number 25198 by Joel578 last updated on 06/Dec/17 $$\mathrm{Given} \\ $$$$\mathrm{sin}\:\left({K}\:+\:{L}\right)\:\mathrm{cos}\:{M}\:=\:\mathrm{2}\:\mathrm{sin}\:{K}\:\mathrm{cos}\:\left({L}\:−\:{M}\right) \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:{M}\:=\:\frac{\mathrm{sin}\:\left({L}\:−\:{K}\right)}{\mathrm{2}\:\mathrm{sin}\:{K}\:\mathrm{sin}\:{L}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com