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@imqwerty @sweetburger
cot (x - pi/2) = cos( x-pi/2) / sin(x - pi/2) cos(a-b) = cos(a)cos(b) +sin(a)sin(b) cos(x-pi/2) = cos(x) cos(pi/2) + sin(x) sin(pi/2) cos(x-pi/2) = cos(x) (0) + sin(x) (1) = sin (x) sin(a-b) = sin(a)cos(b) -cos(a)sin(b) sin(x-pi/2) = sin(x) cos(pi/2) - cos(x) sin(pi/2) sin(x-pi/2) = sin(x) (0) = -cos(x) (1) cos(x-pi/2) / sin(x-pi/2) = sin(x) / [-cos(x)] = -tan(x)
There is an easier way but that is probably the one that you are looking for.
Yes, are you aware of the double angle formula?\[\large\rm tan(\alpha+\beta)= \frac{tan(\alpha)+tan(\beta )}{1-tan(\alpha)tan(\beta )}\]
thank you both! yes, I'm aware
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