Math, asked by boop3CkeenTaniveda, 1 year ago

Evaluate sec41sin49+cos29cosec61-2/√ 3(tan20tan60tan70) 3(sin 2 31+sin 2 59)

Answers

Answered by ARoy
6
[sec41°sin49°+cos29°cosec61°-2/√3(tan20°tan60°tan70°)]/[3(sin²31°+sin²59°)]
=[sec41°sin(90°-41°)+cos29°cosec(90°-29°)-2/√3{tan20°×√3×tan(90°-20°)}]/
[3{sin²31°+sin²(90°-31°)]
=[sec41°cos41°+cos29°sec29°-2(tan20°cot20°)]/[3(sin²31°+cos²31°]
=(1+1-2)/(3×1)
=0
Similar questions