Given Tan α = 3, find (sin α + cos α) & # 178; The speed is anxious, had better have a little step

Given Tan α = 3, find (sin α + cos α) & # 178; The speed is anxious, had better have a little step

tanα=3,sinα=3cosα ,sin²α=9cos²α,sinα=±3√10/10,cosα=±√10/10
(sin α + cos α) &# 178; = 1 + 2Sin α, cos α = 1 + 6 / 10 = 8 / 5, α∈ I or III
(sin α + cos α) ² = 2 / 5, α∈ Ⅱ or Ⅳ

[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]
One wrong word should be:
[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]

Let a + π = x, then the original formula can be reduced to [cosxsin & # 178; x] / [(SiNx / cosx) cos & # 179; (- x)]
Because cos (- x) = cosx, then the original formula = [cosxsin & # 178; x] / [sinxcos & # 178; x] = SiNx / cosx = TaNx,
That is, the original formula = Tan (a + π)

If Tan α = 2, find the value of (sin α + cos α) / (sin α - cos α) + cos & # α

Sixteen fifths