==>
==> tanA + tan2a = 0 OR
The first case is equivalent to , or equivalently, .
This happens only when 3A = 0, or A = 0 degree. (Notice A = 0 does not make the denominator 0, so A = 0 is a valid solution. It can be checked also by using the original equation that A = 0 is indeed a solution.)
The 2nd case implies that
2 - tanA*tan2A = 0. (Notice that this effectively means that the denominator 1 - tanA*tan2A is not equal to 0.)
==> tanA*tan2A = 2
==> <==>.
Now the only value of A that will make the denominator of the last equation 0 is 45 degrees, in which tanA = 1. Thus we exclude A = 45 degrees from the solution and
let , or
, or .
==> or .
The second instance is unacceptable, since it would situate A outside 0<_A<_90°.
Thus we have
.
Therefore the solution set for the original equation is {0, }
