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verify the identity. cot(x-pi/2) = -tan x
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HI!!
hi!
what does "verify" mean in this instance? my guess is it means use the subtraction angle formula for cotangent but i could be wrong
im not sure
i don't know it, so i just looked it up it is \[\cot(A-B)=\frac{\cot(A)\cot(B)+1}{\cot(B)-\cot(A)}\] sort of annoying
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actually now it is easy, since \[\cot(\frac{\pi}{2})=0\]
\[\cot(A-B)=\frac{\cot(A)\cot(B)+1}{\cot(B)-\cot(A)}\] \[\cot(x-\frac{\pi}{2})=\frac{\cot(x)\cot(\frac{\pi}{2})+1}{\cot(\frac{\pi}{2})-\cot(x)}\]
you get \[-\frac{1}{\cot(x)}\] right away
oh ok thanks so much :)
actually I think I'm supposed to show how cot (x-pi/2)= -tan x
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Keep in mind that cot and tan are inverse functions, so: -1/cotx = -tanx
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