Evaluate #(1-costheta)(1+sectheta)=tantheta#?

#(1-costheta)(1+sectheta)=tantheta#

• Let's distribute the terms:

#1+sectheta-costheta-costhetasectheta=tantheta#

• Let's now use #costheta=1/sectheta#:

#1+1/costheta-costheta-costheta1/costheta=tantheta#

#1+1/costheta-costheta-1=tantheta#

#1/costheta-costheta=tantheta#

• Let's combine the left side:

#1/costheta-costheta(costheta/costheta)=tantheta#

#1/costheta-cos^2theta/costheta=tantheta#

#(1-cos^2theta)/costheta=tantheta#

• We can now use the identity: #sin^2theta+cos^2theta=1=>sin^2theta=1-cos^2theta#

#sin^2theta/costheta=sintheta/costheta#

i.e. #sintheta(1-sintheta)=0#

Therefore, either #sintheta=0# or #1-sintheta=0# i.e. #sintheta=1#

But at #sintheta=1#, we have both #tantheta# and #sectheta# as undefined and hence #theta=npi#.

#( 1 - cos t)(1 + 1/(cos t)) = tan t#
#((1 - cos t)(cos t + 1))/cos t = sin t/(cos t)#
#1 - cos^2 t = sin t#
#sin^2 t - sin t = 0#
#sin t( sin t - 1) = 0#
Use unit circle to solve this equation.
a. sin t = 0 --> t = 0, #t= pi#, and #t = 2pi#
b. sin t - 1 = 0 --> sin t = 1 --> #t = pi/2#
General answers:
#t = kpi#
#t = pi/2 + 2kpi#