Evaluate : sec 41.sin 49 +cos 29.cosec 61 - 2/root 3 (tan 2o.tan 60.tan 70) / 3 (sin square 31 + sin square 59)
Answers
Answered by
110
[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
=[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
Answered by
29
Hope it is clear now
Attachments:
Similar questions