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Category: Trigonometry

Question-212457

Question Number 212457 by 281981 last updated on 14/Oct/24 Answered by som(math1967) last updated on 14/Oct/24 $$\:\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow{a}\left({bc}−{a}^{\mathrm{2}} \right)−{b}\left({b}^{\mathrm{2}} −{ca}\right)+{c}\left({ab}−{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}}…

Question-212445

Question Number 212445 by ChantalYah last updated on 13/Oct/24 Commented by Frix last updated on 13/Oct/24 $$\mathrm{80cos}\:{A}\:?\:\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$ \\ $$$$\mathrm{80cos}\:{A}\:−\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$−\mathrm{170sin}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=\mathrm{13} \\…

Determiner-R1-R2-R3-pour-b-12cm-EF-MN-EF-Tangent-aux-cercles-C1-R1-C2-R2-EF-a-MN-b-MN-tangent-au-cercle-C2-OM-ON-3a-2-MON-2x-

Question Number 212011 by a.lgnaoui last updated on 28/Sep/24 $$\boldsymbol{\mathrm{D}}\mathrm{eterminer}:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\boldsymbol{\mathrm{pour}}\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{12}\boldsymbol{\mathrm{cm}}\:\:\:\: \\ $$$$\boldsymbol{\mathrm{EF}}\://\:\boldsymbol{\mathrm{MN}}\:;\:\:\boldsymbol{\mathrm{EF}}\:\boldsymbol{\mathrm{Tangent}}\:\boldsymbol{\mathrm{aux}}\:\boldsymbol{\mathrm{cercles}}:\: \\ $$$$\:\:\boldsymbol{\mathrm{C}}\mathrm{1}\left(\boldsymbol{\mathrm{R}}\mathrm{1}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{2}\left(\boldsymbol{\mathrm{R}}\mathrm{2}\right)\:\:;\:\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{MN}}:\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{cercle}}\:\boldsymbol{\mathrm{C}}\mathrm{2} \\ $$$$\boldsymbol{\mathrm{OM}}=\boldsymbol{\mathrm{ON}}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\measuredangle\mathrm{MON}=\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\: \\ $$$$ \\ $$ Commented…

Question-211956

Question Number 211956 by Durganand last updated on 25/Sep/24 Answered by Frix last updated on 25/Sep/24 $$\mathrm{tan}\:\alpha\:={t}\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha} \\…

In-a-triangle-the-bisector-of-the-side-c-is-perpendicular-to-side-b-Prove-that-2tanC-tanA-0-

Question Number 211672 by MATHEMATICSAM last updated on 15/Sep/24 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{c}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{side}\:{b}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{2tanC}\:+\:\mathrm{tanA}\:=\:\mathrm{0}. \\ $$ Commented by Frix last updated on 16/Sep/24 $$\mathrm{If}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{a}\:{side}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to} \\…