# Math Help - Trigonometry Help

1. ## Trigonometry Help

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?

2. Originally Posted by janvdl
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?
cot (90 + y) = cos (90 + y)/sin (90 + y)

= [cos 90 cos y - sin 90 sin y]/[sin 90 cos y + cos 90 sin y]

= -sin y/cos y

= -tan y

3. Hello, janvdl!

Prove: .cot(90 + θ) .= .- tanθ
. . . . . . . . . . . . . . . . . . . . . . . . . . . .tanα + tanβ
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