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sqrt[3]cosec20 + sec20=?

sqrt[3]cosec20 + sec20=?

Grade:9

3 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 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
10 Points
9 years ago
Really nice methods :)
Suraj
11 Points
6 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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