Question

y=1+sinx-cosx÷1+sinx+cosx then y2= plz send me fast
​

Answer 1

Step-by-step explanation:

y=1+sinx-cosx÷1+sinx+cosx

y=(1+sinx-cosx÷1+sinx+cosx )(1+sinx-cosx÷1+sinx-cosx )

=(1+sinx-cosx)^2 ÷((1+sinx)^2 - (cosx )^2)

=(1+sin^2x+cos^2x+2sinx-2sinxcosx-2cosx)÷(1+sin^2x+2sinx-cos^2x)

=(1+1+2sinx-2sinxcosx-2cosx)÷(sin^2x+sin^2x+2sinx)

since sin^2x+cos^2x=1, 1-cos^2x=sin^2x

=(2+2sinx-2sinxcosx-2cosx)÷(2sin^2x+2sinx)

=(2(1+sinx)-2cosx(1+sinx))÷2sinx(sinx+1)

=2(1+sinx)(1-cosx)÷2sinx(1+sinx)

=(1-cosx)/sinx

y^2=((1-cosx)/sinx)^2

=(1-cosx)^2/sin^2x

=(1-cos)^2/(1-cos^2x )

=(1-cosx)(1-cosx) / ( 1-cosx)(1+cosx)

=(1-cosx)/(1+cosx)