Question

(1+sinα)/（根号下(1+cosα)-根号下(1-cosα))+(1-sinα)/（根号下(1+cosα)+根号下(1-cosα))

已知π<α<3π/2,求(1+sinα)/（根号下(1+cosα)-根号下(1-cosα))+(1-sinα)/（根号下(1+cosα)+根号下(1-cosα))的值

Answer 1

原式=(1+sinα)/√{2[cos(α/2)]^2} - √{2[sin(α/2)]^2} + (1-sinα)/√{2[cos(α/2)]^2} + √{2[sin(α/2)]^2}
∵π < α < 3π/2
∴π/2 < α/2 <3π/4
={[sin(α/2)]^2 + [cos(α/2)]^2 + 2sin(α/2)*cos(α/2)]}/[-√2*cos(α/2) - √2*sin(α/2)]
+ {[cos(α/2)]^2 + [sin(α/2)]^2 - 2sin(α/2)*cos(α/2)]}/[-√2*cos(α/2) + √2*sin(α/2)]
=[sin(α/2) + cos(α/2)]^2 / (-√2)*[cos(α/2) + sin(α/2)] + [cos(α/2) - sin(α/2)]^2 / (-√2)*[cos(α/2) - sin(α/2)]
=-[sin(α/2) + cos(α/2)] / (√2) + {-[cos(α/2) - sin(α/2)]} / (√2)
=-[sin(α/2) + cos(α/2)] / (√2) - [cos(α/2) - sin(α/2)] / (√2)
=-√2*cos(α/2)

Answer 2

原式=(1+sina)[根(1+cosa)+根(1-cosa)]/2cosa+(1-sina)[根(1+cosa)-根(1-cosa)]/2cosa=根(1+cosa)+sina根(1-cosa)/cosa

Answer 3

已知π<α<3π/2,求(1+sinα)/（根号下(1+cosα)-根号下(1-cosα))+(1-sinα)/（根号下(1+cosα)+根号下(1-cosα))的值