Question Number 96792 by me2love2math last updated on 04/Jun/20 Commented by me2love2math last updated on 04/Jun/20 $${question}\:\mathrm{260} \\ $$$$ \\ $$ Commented by mr W…
Question Number 96766 by bemath last updated on 04/Jun/20 $$\mathrm{solve}\::\:\mathrm{tan}\:{x}−\mathrm{tan}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$ Answered by bobhans last updated on 04/Jun/20 $$\Rightarrow\:\mathrm{tan}\:\left({x}\right)−\frac{\mathrm{2tan}\:\left({x}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:\left({x}\right)−\mathrm{tan}\:^{\mathrm{3}} \left({x}\right)−\mathrm{2tan}\:\left({x}\right)=\:\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)…
Question Number 96729 by qwertasdf last updated on 04/Jun/20 $${Prove}\:{that}\:\mathrm{ln}\mid\mathrm{sec}\left({x}\right)+\mathrm{tan}\left({x}\right)\mid=\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{sin}\left({x}\right)\right) \\ $$ Commented by prakash jain last updated on 04/Jun/20 $$\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)=\mathrm{ln}\:\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\mathrm{tanh}^{−\mathrm{1}}…
Question Number 31030 by abdo imad last updated on 02/Mar/18 $$\left.\mathrm{1}\right)\:{solvein}\:{R}\:{the}\:{equation}\:\mathrm{16}{u}^{\mathrm{5}} \:−\mathrm{20}{u}^{\mathrm{3}} \:+\mathrm{5}{u}\:=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{solve}\:{in}\:{R}\:\:{sin}\left(\mathrm{5}{x}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:{sin}\left(\mathrm{5}{x}\right)=\mathrm{16}\:{sin}^{\mathrm{5}} {x}\:−\mathrm{20}\:{sin}^{\mathrm{3}} {x}\:+\mathrm{5}{sinx}\:{by} \\ $$$${using}\:{moivre}\:{formula}. \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{values}\:{of}\:{sin}\left(\frac{\pi}{\mathrm{5}}\right)\:{and}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\left.\mathrm{5}\right)\:{find}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:{and}\:{tan}\left(\frac{\pi}{\mathrm{5}}\right).…
Question Number 31016 by 78987 last updated on 02/Mar/18 $${show}\:{that}\:\mathrm{1}/\mathrm{cosec}{x}−{cotx}+\mathrm{1}/\mathrm{cosec}{x}+\mathrm{cot}{x}=\mathrm{2cosec}{x} \\ $$ Answered by iv@0uja last updated on 02/Mar/18 $$\:\:\:\:\frac{\mathrm{1}}{\mathrm{cosec}\:{x}−\mathrm{cot}\:{x}}+\frac{\mathrm{1}}{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}} \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}+\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}} \\…
Question Number 30929 by rahul 19 last updated on 28/Feb/18 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\:? \\ $$ Commented by abdo imad last updated on 01/Mar/18…
Question Number 161939 by amin96 last updated on 24/Dec/21 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{1010}} {\sum}}\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\frac{\pi\boldsymbol{\mathrm{n}}}{\mathrm{2022}}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30856 by Joel578 last updated on 27/Feb/18 Commented by mrW2 last updated on 28/Feb/18 $${Question}\:{is}\:{wrong}.\:{It}\:{should}\:{be}: \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{A}=… \\ $$ Answered by MJS…
Question Number 161926 by amin96 last updated on 24/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161907 by bobhans last updated on 24/Dec/21 $$\left(\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\frac{\mathrm{27}\pi}{\mathrm{20}}\right)=? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\:\frac{\pi}{\mathrm{30}}\:\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{30}}\:\mathrm{tan}\:\frac{\mathrm{11}\pi}{\mathrm{30}}\:=? \\ $$ Commented by blackmamba last updated on 24/Dec/21 $$\:\frac{\pi}{\mathrm{20}}\:=\:\mathrm{9}°\:;\:{G}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{9}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{27}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{81}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{243}°\right) \\ $$$$\:\left[\:\begin{cases}{\mathrm{cos}\:\mathrm{81}°=\mathrm{sin}\:\mathrm{9}°}\\{\mathrm{cos}\:\mathrm{243}°=−\mathrm{sin}\:\mathrm{27}°\:}\end{cases}\right] \\…