Question Number 128755 by bramlexs22 last updated on 10/Jan/21 $$\mathrm{Solve}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\:+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3x} \\ $$ Answered by liberty last updated on 10/Jan/21 $$''\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}}…
Question Number 128753 by bramlexs22 last updated on 10/Jan/21 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{27}^{\mathrm{cos}\:\mathrm{2x}} .\:\mathrm{81}^{\mathrm{sin}\:\mathrm{2x}} \:? \\ $$ Answered by liberty last updated on 10/Jan/21 $$\:\Leftrightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{3}^{\mathrm{3cos}\:\mathrm{2x}} .\:\mathrm{3}^{\mathrm{4sin}\:\mathrm{2x}}…
Question Number 128743 by bemath last updated on 10/Jan/21 $$\:\:\:\:\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{Find}\:\mathrm{x}\:. \\ $$ Answered by liberty last updated on 10/Jan/21 $$\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{3}}…
Question Number 128681 by bemath last updated on 09/Jan/21 $$\:\mathrm{Given}\:\sqrt[{\mathrm{3}}]{\mathrm{sin}\:\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}} \\ $$$$\:\mathrm{then}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=?\: \\ $$ Commented by MJS_new last updated on 09/Jan/21 $$\mathrm{not}\:\mathrm{funny}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{exactly}\:\mathrm{but}\:\mathrm{possible}… \\ $$…
Question Number 128659 by liberty last updated on 09/Jan/21 $$\:\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{11}}\right)\:+\:\mathrm{4}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{11}}\right)\:=\:?\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128645 by john_santu last updated on 09/Jan/21 $$\:\mathrm{If}\:\mathrm{6tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{tan}\:\mathrm{x}.\mathrm{sec}\:\mathrm{x}\:=\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{where} \\ $$$$\frac{\pi}{\mathrm{2}}<\mathrm{x}<\frac{\mathrm{3}\pi}{\mathrm{2}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} .\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right). \\ $$ Answered by liberty…
Question Number 128641 by john_santu last updated on 09/Jan/21 $$\:\mathrm{Given}\:\begin{cases}{\mathrm{A}+\mathrm{B}=\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{A}−\mathrm{B}=\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{tan}\:\mathrm{A}\:=? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{2A}\:=\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{tan}\:\mathrm{2A}\:=\:\frac{\mathrm{1}/\mathrm{2}+\mathrm{1}/\mathrm{3}}{\mathrm{1}−\left(\mathrm{1}/\mathrm{2}\right)×\left(\mathrm{1}/\mathrm{3}\right)}\:=\:\mathrm{1} \\…
Question Number 128643 by liberty last updated on 09/Jan/21 $$\:\mathrm{If}\:\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}\:=\:\mathrm{b}\:\mathrm{then}\:\mathrm{cos}\:\mathrm{x}\:=? \\ $$ Commented by john_santu last updated on 09/Jan/21 $$\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{b}−\mathrm{sec}\:\mathrm{x}\: \\ $$$$\mathrm{squaring}\:\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}=\mathrm{b}^{\mathrm{2}} −\mathrm{2bsec}\:\mathrm{x}+\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}…
Question Number 63059 by rajesh4661kumar@gamil.com last updated on 28/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63060 by rajesh4661kumar@gamil.com last updated on 28/Jun/19 Answered by Hope last updated on 28/Jun/19 $$\sqrt{\frac{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} }{{sin}^{\mathrm{2}} \theta}}\: \\ $$$$=\mid\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}\mid \\ $$$$=\frac{\mid\mathrm{1}+{cos}\theta\mid}{\mid{sin}\theta\mid} \\ $$$${when}\:\:\:\pi>\theta>\mathrm{0}\:\:{so}\:{sin}\theta=+{ve}…