Question Number 117926 by bemath last updated on 14/Oct/20 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+\:\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2x}}\right)\:=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} +\left(\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} =\frac{\mathrm{5}\pi}{\mathrm{8}} \\ $$$$ \\ $$ Answered…
Question Number 117852 by bemath last updated on 14/Oct/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{24}}+\mathrm{tan}\:\frac{\mathrm{5}\pi}{\mathrm{24}}\right).\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:+\:\mathrm{2}\:. \\ $$$$\left(\mathrm{2}\right)\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5x}}}\:.\left(\mathrm{5}\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{20x}}\:\right)^{\mathrm{0}.\mathrm{5}} .\:\mathrm{2}^{−\mathrm{1}} \:=\:? \\ $$ Answered by bobhans last updated on 14/Oct/20…
Question Number 52082 by ajfour last updated on 03/Jan/19 Commented by ajfour last updated on 03/Jan/19 $${Express}\:\boldsymbol{\alpha}\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$ Answered by mr W last updated…
Question Number 117606 by TANMAY PANACEA last updated on 12/Oct/20 $${find} \\ $$$${sina}+{sin}\left({a}+{b}\right)+{sin}\left({a}+\mathrm{2}{b}\right)++..+{sin}\left\{{a}+\left({n}−\mathrm{1}\right){b}\right\} \\ $$ Answered by Dwaipayan Shikari last updated on 12/Oct/20 $$\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{{b}}{\mathrm{2}}}\left(\mathrm{2}{sinasin}\frac{{b}}{\mathrm{2}}+\mathrm{2}{sin}\left({a}+{b}\right){sin}\frac{{b}}{\mathrm{2}}+…..\mathrm{2}{sin}\left({a}+\left({n}−\mathrm{1}\right){b}\right){sin}\frac{{b}}{\mathrm{2}}\right) \\…
Question Number 117552 by sandy_delta last updated on 12/Oct/20 $$\mathrm{please}\:\mathrm{help} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:.\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:.\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\:=\:? \\ $$ Commented by TANMAY PANACEA last updated on 12/Oct/20 $${find}\:{cos}\frac{\pi}{\mathrm{7}}+{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}+{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}\:\:{pls} \\ $$…
Question Number 183068 by cortano1 last updated on 19/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 117497 by bemath last updated on 12/Oct/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{equation}\:\mathrm{sec}\:\mathrm{3}\theta\:=\:\mathrm{sec}\:\theta \\ $$ Answered by Dwaipayan Shikari last updated on 12/Oct/20 $$\frac{\mathrm{1}}{{cos}\mathrm{3}\theta}=\frac{\mathrm{1}}{{cos}\theta} \\ $$$${cos}\mathrm{3}\theta={cos}\theta…
Question Number 182848 by malithxd last updated on 15/Dec/22 $$\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}} \\…