Prove that (1+sinx-cosx/1+sinx+cosx)^2= 1-cosx/1+cosx

1 answer

For ease of reading, I'll just use s for sinx and c for cosx

(1+s-c)/(1+s+c)
= (1+s-c)^2/[(1+s)^2-c^2)
= (1+s-c-sc)/(s+s^2)
= (1-c)(1+s)/(s(1+s))
= (1-c)/s

That means that

[(1+s-c)/(1+s+c)]^2 = (1-c)^2/s^2
= (1-c)(1-c) / (1+c)(1-c)
= (1-c)/(1+c)
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
    1. answers icon 0 answers
  3. prove the identity(sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2)
    1. answers icon 0 answers
  4. I need to prove that the following is true. Thanks(cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds
    1. answers icon 0 answers
more similar questions