Here's a problem not even my teacher could figure out.
Everyone knows that cot(90 + y) = -tan y
So now we have to prove it.
cot(90 + y) = 1/(tan 90 + y)
LHS = {1 - (tan 90)(tan y)} / {(tan 90)(tan y}
But what the heck are we to do with tan 90???
And how will one prove this identity?
Hello, janvdl!
. . . . . . . . . . . . . . . . . . . . . . . . . . . .tanα + tanβProve: .cot(90 + θ) .= .- tanθ
You know this formula: .tan(α + β) .= .----------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - tanα·tanβ
. . . . . . . . . . . . . . . . . . . . . . . . . . . . cotα·cotβ - 1
We can derive this one: .cot(α + β) .= .----------------
. . . . . . . . . . . . . . . . . . . . . . . . . . . . cotα + cotβ
. . . . . . . . . . . . . . . . cot90·cotθ - 1 . . . 0·cotθ - 1 . . . .-1
Then: .cot(90 + θ) .= .----------------- .= .------------ .= .------ .= .- tanθ
. . . . . . . . . . . . . . . . .cot90 + cotθ . . . .0 + cotθ . . . .cotθ