How do you solve #1=cot^2theta+csctheta# for #0<=theta<=2pi#?

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Answer 1 · 2016-09-20T13:38:40

Apply the identities #cottheta = costheta/sintheta# and #csctheta = 1/sintheta#:

#cos^2theta/sin^2theta + 1/sintheta = 1#

#cos^2theta/sin^2theta + sintheta/sin^2theta = 1#

#(cos^2theta + sin theta)/sin^2theta = 1#

#cos^2theta + sin theta = sin^2theta#

Apply the identity #sin^2theta + cos^2theta = 1 ->cos^2theta = 1 - sin^2theta#

#1 - sin^2theta + sin theta - sin^2theta = 0#

#-2sin^2theta + sin theta + 1 = 0#

#-2sin^2theta + 2sintheta - sin theta + 1 = 0#

#-2sintheta(sin theta - 1) - 1(sin theta - 1) = 0#

#(-2sintheta - 1)(sin theta - 1) = 0#

#sintheta = -1/2 and sin theta = 1#

#theta = (7pi)/6, (11pi)/6 and pi/2#

However, since #pi/2# renders the equation undefined, that solution is extraneous. Hence, our solution set is #{(7pi)/6, (11pi)/6}#.

Hopefully this helps!