Verify and identify, I am so confused on how to do this? this is an extra credit but I am lost, please help.
cot(�Ý-ƒÎ/2)=-tan �Ý
2 answers
I do not understand your symbols, do not have whatever font you are using. I assume this is some kind of a phase shift in the argument of the cotangent.
If you mean
cot(θ-π/2) then that's easy
just as sinθ = cos(π/2-θ),
tanθ = cot(π/2-θ)
and, since cot(-θ) = -cotθ,
cot(θ-π/2) = -cot(π/2-θ) = -tanθ
Oh, well, if you must, show that
cot(θ-π/2) = 1/tan(θ-π/2)
= (1+tanθ*tan π/2)/(tanθ - tanπ/2)
Now divide by tanθ*tan π/2 and you have
= (cotθ*cot π/2+1)/(cotπ/2 - cotθ)
= 1/(0-cotθ)
= -tanθ
cot(θ-π/2) then that's easy
just as sinθ = cos(π/2-θ),
tanθ = cot(π/2-θ)
and, since cot(-θ) = -cotθ,
cot(θ-π/2) = -cot(π/2-θ) = -tanθ
Oh, well, if you must, show that
cot(θ-π/2) = 1/tan(θ-π/2)
= (1+tanθ*tan π/2)/(tanθ - tanπ/2)
Now divide by tanθ*tan π/2 and you have
= (cotθ*cot π/2+1)/(cotπ/2 - cotθ)
= 1/(0-cotθ)
= -tanθ