sin(-19π/6)/(sec4π/3)+tan7π/4=详细过程谢谢这是道选择题A1/6B-1/6C√3/3D-√3/3

Question

sin(-19π/6)/(sec4π/3)+tan7π/4=详细过程谢谢
这是道选择题A1/6B-1/6C√3/3D-√3/3

Answer 1

sin(-19π/6)/(sec4π/3)+tan7π/4=
=sin(-19π/6+2π)/(1/cos(4π/3))+tan3π/4
=sin(-7π/6)×cos(4π/3)-1
=sin(5π/6)cos(4π/3)-1
=1/2×(-1/2)-1
=-1/4-1
=-5/4;
很高兴为您解答,skyhunter002为您答疑解惑
如果本题有什么不明白可以追问,