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Solve tan2A - 2tanA = -1 for 0º≤A≤360º.
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tan^2A - 2tanA = -1*
\[\frac{ 2\tan A }{1-\tan ^{2}A }-2\tan A=-1\] \[2\tan A \left( \frac{ 1 }{1-\tan ^{2}A }-1 \right)=-1\] \[2\tan A \left( \tan ^{2}A \right)=-1+\tan ^{2}A\] \[2\tan ^{3}A=\sec ^{2}A\] R.H.S is positive ,hence L.H.S. is also positive So A lies in ist or 3rd quadrant.
Would the answer be 45 degrees and 225 degrees or 60 and 240?
since both are in the same quadrants...
\[now you have mentioned \it is \tan ^{2}A\]
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yes I'm sorry!
\[\tan ^{2}A-2 \tan A=-1or \tan ^{2}A-2\tan A+1=0\] \[\left( \tan A-1 \right)^{2}=0,\tan A=1,A=45,225\]
Thank you for your assistance!
yw
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