Çözüldü Trigonometri

[ (sin10)^2 ]·[ (1 / cos20)·(1 + 1 / cos40) + 2(cos80 + cos40) / (cos40·cos80) ] = ?
(Tüm sayılar derecedir.)

[ (sin10)^2 ]·{ (1 / cos20)·[ 2(cos20)^2 / cos40 ] + 2(sin10 + cos40) / (cos40·sin10) } =
[ 2(sin10)^2 ]·[ cos20 / cos40 + (sin10 + cos40) / (cos40·sin10) ] =
[ 2(sin10)^2 / cos40 ]·[ (cos20·sin10 + sin10 + cos40) / sin10 ] =
(2sin10 / cos40)·[ sin10·(cos20 + 1) + cos40 ] =
(2sin10 / cos40)·{ sin10·[2(cos10)^2 ] + cos40 } =
(2sin10 / cos40)·(sin20·cos10 + cos40) =
(sin20)^2 / cos40 + 2sin10 =
[ (sin20)^2 + 2sin10·cos40 ] / cos40 =
[ (sin20)^2 + sin50 - sin30 ] / cos40 =
[ 2(sin20)^2 + 2sin50 - 1 ] / (2cos40) =
{ 2sin50 - [ 1 - 2(sin20)^2 ] } / (2cos40) =
(2sin50 - cos40) / (2cos40) =
cos40 / (2cos40) =
1 / 2

WolframAlpha Kontrolu:
https://www.wolframalpha.com/input?i=[ (sin(10°))^2 ]·[ sec(20°)·(1 + sec(40°)) + 2(cos(80°) + cos(40°)) / (cos(40°)·cos(80°)) ] =