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How can arccot of tanx be simplified?
There is not much that can be done by way of simplification. Suppose arccot(y) = tan(x) then y = cot[tan(x)] = 1/tan(tan(x)) Now cot is NOT the inverse of tan, but its reciprocal. So the expression in the first of above equation cannot be simplified further. Similarly tan[tan(x)] is NOT tan(x)*tan(x) = tan2(x)
How do you solve tan squared theta - tan theta equals 0?
Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1
Is tan squared x equal tan x squared?
Yes. Both expressions are the same.
What angle do you need for a gradient of 1 in 20?
If the angle is x radians, then tan(x) = 1/20 = 0.05 So x = tan-1(0.05) = 0.04996 radians = 2.86 degrees (approx)
How do you express cosine in terms of cotangent?
cos(x)=sin(x-tau/4) tan(x)=sin(x)/cos(x) sin(x)=tan(x)*cos(x) cos(x)=tan(x-tau/4)*cos(x-tau/4) you can see that we have some circular reasoning going on, so the best we can do is express it as a combination of sines and cotangents: cos(x)=1/cot(x-tau/4)*sin(x-tau/2) tau=2*pi
