How do you simplify the expression #cot^2x+1#?

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Answer 1 · 2016-09-15T14:10:42

#csc^(2)(x)#

We have: #cot^(2)(x) + 1#

This expression can be simplfiified by applying one of the Pythagorean identities: #1 + cot^(2)(x) = csc^(2)(x)#

#= csc^(2)(x)#

Answer 2 · 2018-07-19T15:24:13

#csc^2x#

Let's say you didn't know the identity: #1+cot^2x= csc^2x#
Another approach could be:

#cot^2x+1=#

#cos^2x/sin^2x+sin^2x/sin^2x=#

#(cos^2x+sin^2x)/sin^2x=#

#1/sin^2x=#

#csc^2x#