Question

evaluate : sec41°sin49°+cos29°sec61°
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Answer 1

Answer:

[sec41°sin49°+cos29°cosec61°-2/√3(tan20°tan60°tan70°)]/[3(sin²31°+sin²59°)]

[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°)}]/

[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°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°]

[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)

[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

SORRY ITS PASTED REPEATEDLY, STILL ......

HOPE IT HELPS ❤️

Answer 2

hyu dude here is ur answer

plz mrk me brainlist