limx→π/4 cot3x - tanxcos(x+π/4)

Let the given limit be L.
L = limx→π/4 (cot3x - tanxcos(x+π/4))
We can rewrite the expression as:
cot3x - tanxcos(x+π/4) = cot3x - tanx(cos x cos(π/4) - sin x sin(π/4))
= cot3x - tanx(cosx(1/√2) - sinx(1/√2))
= cot3x - (tanx/√2)(cosx - sinx)
= cot3x - (sinx/cosx)(cosx - sinx)/(√2)
= cot3x + (sin²x - sinxcosx)/(cosx√2)
As x → π/4, cot3x → cot(3π/4) = -1
Also, as x → π/4, (sin²x - sinxcosx)/(cosx√2) → (1/2 - 1/2)/(1/√2) = 0
Therefore, L = -1 + 0 = -1
However, none of the given options match this result. Let's re-examine the problem.
Let's use L'Hopital's rule. Let f(x) = cot(3x) - tan(x)cos(x+π/4).
Then f(π/4) = cot(3π/4) - tan(π/4)cos(π/2) = -1 - 1(0) = -1.
Let's use the formula cos(A+B) = cosAcosB - sinAsinB
cos(x+π/4) = cosx cos(π/4) - sinx sin(π/4) = (cosx - sinx)/√2
cot3x - tanx((cosx - sinx)/√2) = cot3x - (sinx(cosx - sinx))/(cosx√2)
As x approaches π/4, this expression approaches -1.
Let's check the options again. There must be a mistake in the question or options.
The limit is -1, which is not in the options. There might be an error in the question or options provided.