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}}\: \\ $$$$=\mid\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}\mid \\ $$$$=\frac{\mid\mathrm{1}+{cos}\theta\mid}{\mid{sin}\theta\mid} \\ $$$${when}\:\:\:\pi>\theta>\mathrm{0}\:\:{so}\:{sin}\theta=+{ve}…
Question Number 128582 by john_santu last updated on 08/Jan/21 $$\:\:\mathrm{cot}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cot}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cot}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=? \\ $$ Answered by liberty last updated on 08/Jan/21 $$\:\mathrm{T}\:=\:\mathrm{cot}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cot}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cot}\:^{\mathrm{2}}…
Question Number 128546 by mr W last updated on 08/Jan/21 $${find}\:{a}\:{simple}\:{expression}\:{for} \\ $$$${y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{x}\right) \\ $$$${x}\in\mathbb{R}\:{and}\:{x}\:{in}\:{rad}. \\ $$ Commented by liberty last updated on 08/Jan/21…
Question Number 128361 by DomaPeti last updated on 06/Jan/21 $${f}\left({x}\right)= \\ $$$$\frac{{R}}{\:\sqrt{\mathrm{1}−{M}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({atan}\left({N}\centerdot{sin}\left({x}+{c}\right)\right)\right)}} \\ $$$${g}\left({x}\right)={acos}\left({A}\centerdot{sin}\left({atan}\left({N}\centerdot{sin}\left({x}+{c}\right)\right)\right)+\right. \\ $$$$\left.{B}\centerdot{cos}\left({atan}\left({N}\centerdot{sin}\left({x}+{c}\right)\right)\right)\centerdot{cos}\left({D}−{x}\right)\right) \\ $$$$\int\left({f}\left({x}\right)\centerdot{g}\left({x}\right)'\right)={L}\left({x}\right) \\ $$$${L}\left({x}\right)=? \\ $$$$ \\…
Question Number 128336 by liberty last updated on 06/Jan/21 $$\mathrm{If}\:\:\frac{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}}{\mathrm{sec}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{5}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{3cos}\:\mathrm{2x}+\mathrm{1}}{\mathrm{3cos}\:\mathrm{2x}−\mathrm{1}}\:?\: \\ $$ Answered by bramlexs22 last updated on 06/Jan/21 $$\:\Rightarrow\:\frac{\mathrm{sec}\:{x}+\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}−\mathrm{tan}\:{x}}\:=\:\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}\:=\:\mathrm{5} \\…
Question Number 128333 by liberty last updated on 06/Jan/21 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:\mathrm{30}°\mathrm{30}'\:=\:\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:−\mathrm{2} \\ $$ Commented by MJS_new last updated on 06/Jan/21 $$\mathrm{not}\:\mathrm{true}?! \\ $$$$\mathrm{tan}\:\left(\mathrm{7}.\mathrm{5}°+\mathrm{180}°×{n}\right)\:=−\mathrm{2}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$ Commented…
Question Number 62761 by lalitchand last updated on 25/Jun/19 $$\mathrm{if}\:\mathrm{cos}\theta=\frac{\mathrm{cosA}.\mathrm{cosB}}{\mathrm{1}−\mathrm{cosA}.\mathrm{cosB}}\:\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{Tan}\frac{\theta}{\mathrm{2}}=\mathrm{Tan}\frac{\mathrm{A}}{\mathrm{2}}.\mathrm{Cot}\frac{\mathrm{B}}{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 25/Jun/19 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{cosA}−\mathrm{cosB}}{\mathrm{1}−\mathrm{cosAcosB}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\mathrm{1}−\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosAcosB}}{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}−\mathrm{cosAcosB}}…
Question Number 62698 by Ankit0512 last updated on 24/Jun/19 $${Lines}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{10}=\mathrm{0}\:{and}\:\mathrm{5}{x}−\mathrm{12}{y}−\mathrm{40}=\mathrm{0} \\ $$$${touch}\:{circle}\:{C}_{\mathrm{1}} \:{of}\:{diameter}\:\mathrm{6}.\:{If}\:{the}\: \\ $$$${center}\:{of}\:{C}_{\mathrm{1}} \:{lies}\:{in}\:{the}\:{Ist}\:{quadrant}, \\ $$$${find}\:{the}\:{equation}\:{of}\:{circle}\:{C}_{\mathrm{2}} \:{which}\:{is}\: \\ $$$${concentric}\:{with}\:{C}_{\mathrm{1}\:} \:{and}\:{cuts}\:{intercept} \\ $$$${of}\:{length}\:\mathrm{8}\:{on}\:{these}\:{lines}. \\…