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 $$\:\frac{\pi}{\mathrm{20}}\:=\:\mathrm{9}°\:;\:{G}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{9}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{27}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{81}°\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{243}°\right) \\ $$$$\:\left[\:\begin{cases}{\mathrm{cos}\:\mathrm{81}°=\mathrm{sin}\:\mathrm{9}°}\\{\mathrm{cos}\:\mathrm{243}°=−\mathrm{sin}\:\mathrm{27}°\:}\end{cases}\right] \\…
Question Number 161903 by amin96 last updated on 23/Dec/21 $$\frac{\mathrm{1}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}=??? \\ $$$$ \\ $$ Commented by blackmamba last updated on 24/Dec/21 $$\:=\frac{\mathrm{1}−\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{1}+\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\…
Question Number 30823 by drishtantsingh315 last updated on 26/Feb/18 Answered by math1967 last updated on 27/Feb/18 $${Cos}\left[\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{sin}\:\left(\mathrm{cot}^{−\mathrm{1}} {x}\right)\right\}\right] \\ $$$$\mathrm{cos}\:\left[\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}\right)\right\}\right] \\…
Question Number 96282 by 675480065 last updated on 31/May/20 Answered by bemath last updated on 31/May/20 $$\mathrm{look}\:\mathrm{qn}\:\mathrm{94383} \\ $$ Answered by 1549442205 last updated on…
Question Number 96260 by bemath last updated on 31/May/20 $$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{7}\pi}{\mathrm{9}}\right)\:=? \\ $$ Answered by john santu last updated on 31/May/20 $$\mathrm{set}\:{x}\:=\:\frac{\pi}{\mathrm{9}},\:\frac{\mathrm{4}\pi}{\mathrm{9}},\:\frac{\mathrm{7}\pi}{\mathrm{9}} \\ $$$$\mathrm{3}{x}\:=\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{7}\pi}{\mathrm{3}}\: \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:=\:\mathrm{tan}\:\frac{\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{4}\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{3}}\:=\:\sqrt{\mathrm{3}}…