Category: Trigonometry

Question Number 174556 by mnjuly1970 last updated on 04/Aug/22 Commented by infinityaction last updated on 04/Aug/22 $$\tan \alpha +3=2\sec \alpha$$$${3\tan }^2\alpha -6\tan \alpha -5=0$$$$\tan \alpha =\frac{6\pm 4\sqrt{6}}{6}=\frac{\sin \alpha }{\cos \alpha }$$$$\:\:\:\:\:\:\:\:\frac{\mathrm{sin}\alpha−\mathrm{cos}\alpha\:\:}{\mathrm{sin}\alpha+\mathrm{cos}\alpha\:\:}\:\:=\:\:\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{12}+\mathrm{4}\sqrt{\mathrm{6}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \…

Question Number 174435 by cortano1 last updated on 01/Aug/22 $$\sec {}^{2}1°+\sec {}^{2}2°+\sec {}^{2}3°+\dots +\sec {}^{2}89°=?$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 174346 by cortano1 last updated on 30/Jul/22 $$\frac{1}{\cos {}^{2}10°}+\frac{1}{\sin {}^{2}20°}+\frac{1}{\sin {}^{2}40°}-\frac{1}{\cos {}^{2}45°}=?$$ Answered by behi834171 last updated on 30/Jul/22 $$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{10}}=\mathrm{1}+{tg}^{\mathrm{2}}…

Question Number 43267 by Rauny last updated on 09/Sep/18 $$\cos 2x={2\cos }^2-1$$$$\cos 3x={4\cos }^{3}x-3\cos x$$$$\cos 4x={8\cos }^{4}x-{8\cos }^{2}x+1$$$$\cos 5x={16\cos }^{5}x-{20\cos }^3+5\cos x$$$$\mathrm{cos}\:\mathrm{6}{x}=\mathrm{32cos}^{\mathrm{6}} \:{x}−\mathrm{48cos}^{\mathrm{4}}…

Question Number 108780 by 150505R last updated on 19/Aug/20 Answered by bobhans last updated on 19/Aug/20 $$\cot {}^{2}81°=\tan {}^{2}9$$$$\cot {}^{2}63°=\tan {}^{2}27$$$$\left(\ast\right)\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cot}\:^{\mathrm{2}}…