Question Number 140494 by benjo_mathlover last updated on 08/May/21 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)+\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}\right)=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{x}=? \\ $$ Answered by john_santu last updated on 08/May/21 $$\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{cot}^{−\mathrm{1}}…
Question Number 9414 by Rohit kumar last updated on 06/Dec/16 $${tanA}+{cot}={A}=\:\mathrm{2}{cosec}\mathrm{2}{A} \\ $$ Answered by ridwan balatif last updated on 06/Dec/16 $$\mathrm{tanA}+\mathrm{cotA}=\mathrm{2cosec2A} \\ $$$$\frac{\mathrm{sinA}}{\mathrm{cosA}}+\frac{\mathrm{cosA}}{\mathrm{sinA}}=\mathrm{2cosec2A} \\…
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Question Number 9370 by tawakalitu last updated on 03/Dec/16 Answered by geovane10math last updated on 03/Dec/16 $$\mathrm{There}\:\mathrm{are}\:\mathrm{infinite}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sizes}\:\mathrm{10} \\ $$$$\mathrm{and}\:\mathrm{15}.\:\mathrm{It}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{inclination}\:\mathrm{of}\:{X}\hat {{Z}M}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{w}\:\mathrm{depends}\:\mathrm{of}\: \\ $$$${X}\hat {{Z}M}.…
Question Number 74853 by Raxreedoroid last updated on 01/Dec/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{combine}\:\mathrm{2}{cos}\left(\mathrm{90}°{x}\right)+{cos}\left(\mathrm{180}°{x}\right) \\ $$$$\mathrm{into}\:\mathrm{a}\:\mathrm{form}\:\mathrm{of}\:\:{a}\centerdot{cos}\left({b}\centerdot{x}+{c}\right) \\ $$$$\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}{x}\right)+{cos}\left(\pi{x}\right)\overset{?} {=}{a}\centerdot{cos}\left({bx}+{c}\right) \\ $$ Commented by mr W last updated on 01/Dec/19…
Question Number 140382 by john_santu last updated on 07/May/21 $$\mathrm{sin}\:^{\mathrm{10}} {x}\:+\:\mathrm{cos}\:^{\mathrm{10}} {x}\:=\:\frac{\mathrm{61}}{\mathrm{256}} \\ $$ Answered by MJS_new last updated on 07/May/21 $$\mathrm{sin}\:{x}\:={s}\wedge\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\mathrm{5}{s}^{\mathrm{8}}…
Question Number 140378 by benjo_mathlover last updated on 07/May/21 $$\sqrt{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}}−\sqrt{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right] \\ $$ Commented by john_santu last updated on 07/May/21…
Question Number 140368 by meetbhavsar25 last updated on 06/May/21 Answered by benjo_mathlover last updated on 07/May/21 $$\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\right)\:=\:\mathrm{u} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\:=\:\mathrm{cos}\:\mathrm{u} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{5cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}+\mathrm{3cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{5}−\mathrm{3cos}\:\mathrm{u}\right)\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}−\mathrm{3} \\…
Question Number 140338 by Arzoon last updated on 06/May/21 $${if}\:\theta=\frac{\pi}{\mathrm{2}{n}+\mathrm{1}},\:{n}\geqslant\mathrm{1},\:{n}\in{N}\:,{then}\:{prove}\:{that}: \\ $$$${tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:{tan}\left({n}\theta\right)\:=\:\sqrt{\mathrm{2}{n}+\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9211 by tawakalitu last updated on 23/Nov/16 $$\mathrm{If} \\ $$$$\mathrm{sinA}\:+\:\mathrm{sinB}\:=\:\mathrm{a} \\ $$$$\mathrm{tanA}\:+\:\mathrm{tanB}\:=\:\mathrm{b} \\ $$$$\mathrm{secA}\:+\:\mathrm{secB}\:=\:\mathrm{c} \\ $$$$\mathrm{Prove}\:\mathrm{that}, \\ $$$$\mathrm{8bc}\:=\:\mathrm{a}\left[\mathrm{4b}^{\mathrm{2}} \:+\:\left(\mathrm{b}^{\mathrm{2}} \:−\:\mathrm{c}^{\mathrm{2}} \right)\right]^{\mathrm{2}} \\ $$…