How do you solve #tantheta+cottheta=2# in the interval #0 ≤ x ≤ 2pi#?

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Answer 1 · 2016-11-15T20:02:18

#theta = pi/4#

#sintheta/costheta + costheta/sintheta = 2#

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

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

Recall the identities #sin^2beta + cos^2beta = 1# and #2cosbetasinbeta = sin2beta#.

#1 = sin2theta#

#pi/2 = 2theta#

#theta = pi/4#

Hopefully this helps!