How do you find the exact value of #csc^2theta-cottheta-1=0# in the interval #0<=theta<360#?

1 Answer
Dec 13, 2016

#45^@, 90^@, 225^@, 270^@#

Explanation:

#csc^2 t - cot t - 1 = 0#
#1/(sin^2 t) - cos t/(sin t) = 1#
Use common denominator then cross multiply
#1 - cos t.sin t = sin^2 t#
#1 - sin^2 t - cos t.sin t = 0#
Replace #(1 - sin^2 t)# by #cos^2 t# -->
#cos^2 t - sin t cos t) = 0#
cos t(cos t - sin t) = 0

a. cos t = 0 --> 2 solution arcs -->
arc# t = 90^@# and arc #t = 270^@#
b . cos t - sin t = - (sin t - cos t) = 0
Use trig identity:
#sin a - cos a = - sqrt2cos (a + 45)#
Replace a by t -->
#cos t - sin t = - (sin t - cos t) = sqrt2cos (t + 45) = 0#
cos (t + 45) = 0 --> 2 solution arcs -->
c. t + 45 = 90 --> #t = 90 - 45 = 45^@#
d. t + 45 = 270 --> #t = 270 - 45 = 225^@#
Answers for (0, 360):
#45^@, 90^@, 225^@, 270^@#