How do you solve #(1+tan theta)/(1-tan theta) = (1-tan theta)/(1+tan theta)# ?

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Answer 1 · 2017-01-21T21:37:08

#theta = npi# for any integer #n#

Let #t = tan theta#

Then we want to solve:

#(1+t)/(1-t) = (1-t)/(1+t)#

Multiply both sides by #(1-t)(1+t)# to get:

#(1+t)^2 = (1-t)^2#

Expand to get:

#1+2t+t^2 = 1-2t+t^2#

Subtract #1+t^2# from both sides to get:

#2t = -2t#

Add #2t# to both sides to get:

#4t = 0#

Hence:

#t = 0#

So it is necessary and sufficient that #tan theta = 0#

Note that #tan 0 = 0# and #tan# has period #pi#

So there are solutions::

#theta = npi# for any integer #n#