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Grade 9Trigonometry

sqrt[3]cosec20 + sec20=?

Profile image of Leo Pixie
11 Years agoGrade 9
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3 Answers

Profile image of Arun Kumar
11 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
Profile image of Leo Pixie
11 Years ago
Really nice methods :)
Profile image of Suraj
8 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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