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5cos-2-pi-3-4sec-2-pi-6-tan-2-pi-4-sin-2-pi-6-cos-2-pi-6-easy-mode-

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Question Number 222025 by fantastic last updated on 15/Jun/25
(((5cos^2 (π/3)+4sec^2 (π/6)−tan^2 (π/4))/(sin^2 (π/6)+cos^2 (π/6))))=?? [easy mode]
$$\left(\frac{\mathrm{5cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{3}}+\mathrm{4sec}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}−\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{4}}}{\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}+\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}}\right)=?? \\ $$$$\left[{easy}\:{mode}\right] \\ $$
Answered by MrGaster last updated on 16/Jun/25
cos(π/3)=(1/2),cos^2 (π/3)=((1/2))^2 =(1/4) cos(π/6)=((√3)/2),sec(π/6)=(1/(cos(π/6)))=(1/((√3)/2))=(2/( (√3))),sec^2 (π/6)=((2/( (√3))))^2 =(4/3) tan(π/4)=1,tan^2 (π/4)=1^2 =1 Molecule: 5∙cos^2 (π/3)+4∙sec^2 (π/6)−tan^2 (π/4)=5∙(1/4)+4∙(4/3)−1=(5/4)+((16)/3)−1 General score (denominator 12): (5/4)=((15)/(12)),((16)/3)=((64)/(12)),1=((12)/(12)) ((15)/(12))+((64)/(12))−((12)/(12))=((15+64−12)/(12))=((67)/(12)) Denominator: sin^2 (π/6)+cos^2 (π/6)=((1/2))^2 +(((√3)/2))^2 =(1/4)+(3/4)=1 (∵∀θ,sin^2 θ+cos^2 θ=1) (((67)/(12))/1)=((67)/(12))
$$\mathrm{cos}\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{3}}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{cos}\frac{\pi}{\mathrm{6}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}},\mathrm{sec}\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{cos}\frac{\pi}{\mathrm{6}}}=\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}},\mathrm{sec}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{tan}\frac{\pi}{\mathrm{4}}=\mathrm{1},\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}=\mathrm{1}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{Molecule}: \\ $$$$\mathrm{5}\centerdot\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{3}}+\mathrm{4}\centerdot\mathrm{sec}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}−\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}=\mathrm{5}\centerdot\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{4}\centerdot\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{16}}{\mathrm{3}}−\mathrm{1} \\ $$$$\mathrm{General}\:\mathrm{score}\:\left(\mathrm{denominator}\right. \\ $$$$\left.\mathrm{12}\right): \\ $$$$\frac{\mathrm{5}}{\mathrm{4}}=\frac{\mathrm{15}}{\mathrm{12}},\frac{\mathrm{16}}{\mathrm{3}}=\frac{\mathrm{64}}{\mathrm{12}},\mathrm{1}=\frac{\mathrm{12}}{\mathrm{12}} \\ $$$$\frac{\mathrm{15}}{\mathrm{12}}+\frac{\mathrm{64}}{\mathrm{12}}−\frac{\mathrm{12}}{\mathrm{12}}=\frac{\mathrm{15}+\mathrm{64}−\mathrm{12}}{\mathrm{12}}=\frac{\mathrm{67}}{\mathrm{12}} \\ $$$$\mathrm{Denominator}: \\ $$$$\mathrm{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}+\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{1}\:\left(\because\forall\theta,\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{1}\right) \\ $$$$\frac{\frac{\mathrm{67}}{\mathrm{12}}}{\mathrm{1}}=\frac{\mathrm{67}}{\mathrm{12}} \\ $$

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