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sqrt[3]cosec20 + sec20=?

Leo Pixie , 10 Years ago
Grade 9
anser 3 Answers
Arun Kumar

Last Activity: 10 Years ago

Hello Student,

\\\sqrt3cosec20 + sec20 \\={\sqrt3 \over sin20}+{1 \over cos20} \\={\sqrt3cos20+sin20 \over {1 \over 2}sin40} \\=>{2cos50 \over {1 \over 2}sin40} \\=>4

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Leo Pixie

Last Activity: 10 Years ago

Really nice methods :)

Suraj

Last Activity: 7 Years ago

=√3/sin20°-1/cos20°=2(√3cos20°-sin20°)/2(sin20° cos20°)=2(1/2(√3cos20°-sin20°))/(1/2) sin40°=4(sin60°cos20°-cos60°sin20°)/sin40°=4(sin(60°-20°)/sin40°=4

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