Question

三道高一数学题，求高人指点，都是化简的…

一、根号下1+sinX减根号下1-sinX其中X属于<0~派>
二、f(X)=2tanX-(2sin的方*X/2-1)/(sinX/2*cos/2)
三、[1/tan(X/2)-tan(X/2)]*(1+tan的方X)

Answer 1

①√(1+sinx)-√(1-sinx)
=√[√(1+sinx)-√(1-sinx)]^2
=√[1+sinx+1-sinx-2√(1-sin^2x)]
=√(2-2cosx)

②f(x)=2tanx-[2sin^2(x/2)-1]/(sinx/2*cosx/2)
=2tanx-cosx/(1/2sinx)
=2tanx-2cosx/sinx
=2sinx/cosx-2cosx/sinx
=2[(sin^2x-cos^2x)/sinx*cosx]
=2(-cos2x)/(1/2sin2x)
=-4cot2x

③[1/tan(x/2-tan(x/2]*(1+tan^2x)
=[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]*(1+tan^2x)
=[(cos^2x-sin^2x)/sinxcosx]*(1+tan^2x)
=[cosx/(1/2sinx)]*(1+tan^2x)
=2cotx*(1+tan^2x)
=2cotx+2tanx
=2(cosx/sinx+sinx/cosx)
=2(cos^2x+sin^x)/sinxcosx
=2(cos^2-1/2+sin^2-1/2+1)/(1/2sin2x)
=4(1/2cos2x-1/2cos2x+1)/(sin2x)
=4/sin2x