Category: Trigonometry

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

Question Number 212384 by a.lgnaoui last updated on 14/Oct/24 $$\mathrm{Reponse}\mathrm{a}\mathrm{la}\mathrm{question}\mathrm{N}°$$$$\mathrm{Q}212291\mathrm{S}\left(\mathrm{ABCD}\right)=66,69$$ Commented by hardmath last updated on 12/Oct/24 $$\mathrm{dear}\mathrm{professor},\mathrm{answer}:90$$ Terms…

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

Question Number 211956 by Durganand last updated on 25/Sep/24 Answered by Frix last updated on 25/Sep/24 $$\tan \alpha =t\tan 2\alpha =\frac{2t}{1-t^2}\sin 2\alpha =\frac{2t}{1+t^2}$$$$\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} \…

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