How do you solve: #cot(-x)cot(π/2-x)#?

1 Answer
Douglas K.
Apr 27, 2018

Use the identity #cot(-x) = -cot(x)#:

#cot(-x)cot(π/2-x) = -cot(x)cot(π/2-x)#

Use the identity #cot(pi/2-x) = tan(x)#:

#cot(-x)cot(π/2-x) = -cot(x)tan(x)#

Use the identity #cot(x)tan(x) = 1#:

#cot(-x)cot(π/2-x) = -1, x in RR#

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