求1/cos50°+tan10°的值，要过程，快，谢谢
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1/cos50°+tg10°=1/sin40° + sin10°/cos10°
=1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°)
=1/sin40° +2(sin10°)^2 /sin20°
=1/sin40° +(1-cos20°) /sin20°
=1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20°
=[1+2cos20°-2(cos20°)^2] /sin40°
=(2cos20°-cos40°) /sin40°
=(cos20°+2sin30°sin10°) /sin40°
=(sin70°+sin10°) /sin40°
=[sin(40°+30°)+sin(40°-30°)] /sin40°
=2sin40°cos30°/sin40°=2cos30°=√3
=1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°)
=1/sin40° +2(sin10°)^2 /sin20°
=1/sin40° +(1-cos20°) /sin20°
=1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20°
=[1+2cos20°-2(cos20°)^2] /sin40°
=(2cos20°-cos40°) /sin40°
=(cos20°+2sin30°sin10°) /sin40°
=(sin70°+sin10°) /sin40°
=[sin(40°+30°)+sin(40°-30°)] /sin40°
=2sin40°cos30°/sin40°=2cos30°=√3
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