How do you simplify # (cot^2 x)/(csc x +1)#?

1 Answer
Truong-Son N.
May 5, 2016

The identity you're looking for is #csc^2x - 1 = cot^2x#.

You can derive this by starting from #sin^2x + cos^2x = 1#:

#sin^2x + cos^2x = 1#

#cancel(sin^2x/sin^2x)^(1) + stackrel(cot^2x)overbrace(cos^2x/sin^2x) = stackrel(csc^2x)overbrace(1/sin^2x)#

#\mathbf(1 + cot^2x = csc^2x)#

Thus:

#color(blue)(cot^2x/(cscx + 1))#

#= (csc^2x - 1)/(cscx + 1)#

#= ((cscx - 1)cancel((cscx + 1)))/cancel((cscx + 1))#

#= color(blue)(cscx - 1)#

...if and only if #cscx + 1 ne 0#.

If #cscx + 1 = 0#, then #1/sinx = - 1#, which is the case when #x = (3pi)/2 pm 2npi# for all #n in ZZ#.

Therefore, this answer is valid when #x ne (3pi)/2 pm 2npi# for all #n in ZZ#.

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