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Question Number 177964 by mathlove last updated on 11/Oct/22

(1/(sec15 sin15 cos30))=?

$$\frac{\mathrm{1}}{{sec}\mathrm{15}\:{sin}\mathrm{15}\:{cos}\mathrm{30}}=? \\ $$

Commented by cortano1 last updated on 11/Oct/22

((cos 15°)/((1/2)(√3) sin 15°)) = (2/( (√3))) cot 15°=(2/( (√3)))×(2+(√3)) = ((4+2(√3))/( (√3))) =2+((4(√3))/3) (•) tan 15°=((1−(1/( (√3))))/(1+(1/( (√3))))) ⇒cot 15°=(((√3)+1)/( (√3)−1)) = (1/2)(4+2(√3))=2+(√3)

$$\:\frac{\mathrm{cos}\:\mathrm{15}°}{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{15}°}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cot}\:\mathrm{15}°=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\:=\mathrm{2}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\left(\bullet\right)\:\mathrm{tan}\:\mathrm{15}°=\frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\mathrm{cot}\:\mathrm{15}°=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right)=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 11/Oct/22

(1/(sec15°sin15°cos30°))=((cos15°)/(sin15°))×(1/(cos30°))=(2/( (√3)tan15°)) cos2x=2cos^2 x−1=(2/(sec^2 x))−1=(2/(1+tan^2 x))−1 ⇒cos30°=(2/(1+tan^2 15°))−1 , x=15° ⇒((√3)/2)+1=(2/(1+tan^2 15°)) ⇒(((√3)+2)/2)=(2/(1+tan^2 15°)) ⇒1+tan^2 15°=(4/( (√3)+2))=4(2−(√3)) ⇒tan15°=(√(7−4(√3)))=(√(((√3)−(√4))^2 ))=(√4)−(√3) ⇒(1/(sec15°sin15°cos30°))=(2/( (√3)((√4)−(√3))))=((2(2+(√3)))/( (√3)))

$$\frac{\mathrm{1}}{\mathrm{sec15}°\mathrm{sin15}°\mathrm{cos30}°}=\frac{\mathrm{cos15}°}{\mathrm{sin15}°}×\frac{\mathrm{1}}{\mathrm{cos30}°}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\mathrm{tan15}°} \\ $$$$\mathrm{cos2}{x}=\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{sec}^{\mathrm{2}} {x}}−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos30}°=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}−\mathrm{1}\:,\:{x}=\mathrm{15}° \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°}\:\Rightarrow\frac{\sqrt{\mathrm{3}}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}°=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}+\mathrm{2}}=\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\: \\ $$$$\Rightarrow\mathrm{tan15}°=\sqrt{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}=\sqrt{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{4}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sec15}°\mathrm{sin15}°\mathrm{cos30}°}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by mathlove last updated on 12/Oct/22

thanks

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 12/Oct/22

(1/(sec15sin15cos30 ))=? =(1/(sec15sin15sin60_(▼) )) sin60=2sin30cos30=2(2sin15cos15)cos30 =4sin15cos15cos30 =(1/(sec15sin15∙4sin15cos15cos30 )) =(1/(4sin^2 15_(▼) cos30 )) sin^2 α=((1−cos2α )/2)⇒sin^2 15=((1−cos30 )/2) =(1/(4(((1−cos30 )/2))cos30 )) =(1/(2cos30(1−cos30)))=(1/(2((((√3) )/2))(1−(((√3) )/2)))) =(2/(2(√3) −3))∙((2(√3) +3)/(2(√3) +3))=((4(√3) +6)/3) Formulas Used: •sin2α=2sinαcosα determinant ((( determinant (((•sin^2 α=((1−cos2α )/2)_( _( •sin2α=2sinαcosα) ) ^(Formulas Used:_ ) )))_ ^ )))

$$\frac{\mathrm{1}}{\mathrm{sec15sin15cos30}\:\:\:}=? \\ $$$$=\frac{\mathrm{1}}{\mathrm{sec15sin15}\underline{\underset{\blacktrinagledown} {\mathrm{sin60}}}\:\:\:} \\ $$$$\:\:\mathrm{sin60}=\mathrm{2sin30cos30}=\mathrm{2}\left(\mathrm{2sin15cos15}\right)\mathrm{cos30} \\ $$$$\:\:\:\:\:=\mathrm{4sin15cos15cos30}\: \\ $$$$=\frac{\mathrm{1}}{\cancel{\mathrm{sec15}sin15}\centerdot\mathrm{4sin15}\cancel{\mathrm{cos15}cos30}\:\:\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\underline{\underset{\blacktrinagledown} {\mathrm{sin}^{\mathrm{2}} \mathrm{15}}cos30}\:\:\:} \\ $$$$\:\:\mathrm{sin}^{\mathrm{2}} \alpha=\frac{\mathrm{1}−\mathrm{cos2}\alpha\:}{\mathrm{2}}\Rightarrow\mathrm{sin}^{\mathrm{2}} \mathrm{15}=\frac{\mathrm{1}−\mathrm{cos30}\:}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{1}−\mathrm{cos30}\:}{\mathrm{2}}\right)\mathrm{cos30}\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2cos30}\left(\mathrm{1}−\mathrm{cos30}\right)}=\frac{\mathrm{1}}{\cancel{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}\:}{\cancel{\mathrm{2}}}\right)\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{3}}\centerdot\frac{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{3}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}\:+\mathrm{6}}{\mathrm{3}} \\ $$$$\mathcal{F}{ormulas}\:\:\mathcal{U}{sed}:\:\:\bullet\mathrm{sin2}\alpha=\mathrm{2sin}\alpha\mathrm{cos}\alpha\:\:\: \\ $$$$\begin{array}{|c|}{\:\begin{array}{|c|}{\underset{\:\underset{\:\:\:\:\:\:\:\:\:\bullet\mathrm{sin2}\alpha=\mathrm{2sin}\alpha\mathrm{cos}\alpha} {\:}} {\bullet\mathrm{sin}^{\mathrm{2}} \alpha=\frac{\mathrm{1}−\mathrm{cos2}\alpha\:}{\mathrm{2}}}}\\\hline\end{array}_{\:} ^{\:} }\\\hline\end{array} \\ $$

Commented by Tawa11 last updated on 11/Oct/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Commented by mathlove last updated on 12/Oct/22

thanks sir

$${thanks}\:{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 12/Oct/22

(1/(sec15sin15cos30 ))=? =(1/(sec15sec15cos15 sin15cos30 )) [∵sec15cos15=1] =(1/(sec^2 15∙(1/2)sin30 cos30 )) [∵ cosα sinα=(1/2)sin2α] =(1/((2/(1+cos30 ))∙(1/2)sin30 cos30 )) [∵ sec^2 α=(2/(1+cos2α ))] =(1/((1/(1+cos30 ))∙(1/2)sin60 )) =2(1+cos30)∙(1/(sin60)) ={2+2((((√3) )/2))}.(2/( (√3)))=((4+2(√3))/( (√3)))∙(((√3) )/( (√3)))=((6+4(√3) )/3)

$$\frac{\mathrm{1}}{\mathrm{sec15sin15cos30}\:\:\:}=? \\ $$$$=\frac{\mathrm{1}}{\mathrm{sec15sec15}\underline{\mathrm{cos15}\:\mathrm{sin15}cos30}\:\:\:}\:\left[\because\mathrm{sec15cos15}=\mathrm{1}\right]\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \mathrm{15}\centerdot\underline{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin30}}\:\mathrm{cos30}\:\:\:}\:\left[\because\:\mathrm{cos}\alpha\:\mathrm{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}\alpha\right]\: \\ $$$$=\frac{\mathrm{1}}{\frac{\cancel{\mathrm{2}}}{\mathrm{1}+\mathrm{cos30}\:}\centerdot\frac{\mathrm{1}}{\cancel{\mathrm{2}}}\mathrm{sin30}\:\mathrm{cos30}\:\:\:}\:\left[\because\:\mathrm{sec}^{\mathrm{2}} \alpha=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos2}\alpha\:}\right]\: \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos30}\:}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin60}\:\:\:} \\ $$$$=\mathrm{2}\left(\mathrm{1}+\mathrm{cos30}\right)\centerdot\frac{\mathrm{1}}{\mathrm{sin60}}\: \\ $$$$=\left\{\mathrm{2}+\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\right)\right\}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\sqrt{\mathrm{3}}\:}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{3}}\:}{\mathrm{3}} \\ $$

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