已知tan(180+α)-tan(450-α)=2（0＜α＜90）求[sec(360+α）-sin(450-α)]/[csc(360-α）-cos(180-α)]的值

Question

已知tan(180+α)-tan(450-α)=2（0＜α＜90）求[sec(360+α）-sin(450-α)]/[csc(360-α）-cos(180-α)]的值

Answer 1

∵tan(180°+α)-tan(450°-α)=2 ==>tanα-cotα=2
==>sinα/cosα-cosα/sinα=2
==>sin²α-cos²α=2sincαosα
==>-cos(2α)=sin(2α)
==>tan(2α)=-1
==>2α=135° (∵0＜α＜90°)
==>sin(2α)=1/√2,cos(2α)=-1/√2
∴[sec(360°+α）-sin(450°-α)]/[csc(360°-α）-cos(180°-α)]
=(secα-cosα)/(-cscα+cosα)
=(1/cosα-cosα)/(-1/sinα+cosα)
=[sinα(1-cos²α)]/[cosα(sinαcosα-1)]
=[(sinα)^4]/[sinαcosα(sinαcosα-1)] (分子分母同乘sinα)
=[(1-cos(2α))/2]²/[(sin(2α)/2)(sin(2α)/2-1)] (应用倍角公式)
=[(1+1/√2)/2]²/[((1/√2)/2)((1/√2)/2-1)]
=-(11+8√2)/7.