solution: The area of triangle $ABC$ is given by
  \[\frac{1}{2} bc \sin A.\]Hence,
  \[\frac{1}{2} bc \sin A = a^2 - (b - c)^2 = a^2 - b^2 + 2bc - c^2.\]By the Law of Cosines, $b^2 + c^2 - 2bc \cos A = a^2,$ so
  \[\frac{1}{2} bc \sin A = 2bc - 2bc \cos A.\]This simplifies to $\sin A = 4 - 4 \cos A.$  Squaring both sides, we get
  \[\sin^2 A = 16 - 32 \cos A + 16 \cos^2 A,\]so $1 - \cos^2 A = 16 - 32 \cos A + 16 \cos^2 A.$  This simplifies to
  \[17 \cos^2 A - 32 \cos A + 15 = 0.\]This factors as $(\cos A - 1)(17 \cos A - 15) = 0.$  Since $\cos A$ cannot be equal to 1, $\cos A = \frac{15}{17}.$
  
  Then $\sin A = 4 - 4 \cos A = \frac{8}{17},$ so
  \[\tan A = \frac{\sin A}{\cos A} = \boxed{\frac{8}{15}}.\]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  The area of triangle $ABC$ is equal to $a^2 - (b - c)^2,$ where $a,$ $b,$ and $c$ are the sides of triangle $ABC,$ as usual.  Compute $\tan A.$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \frac{8}{15}
extra_info.question: The area of triangle $ABC$ is equal to $a^2 - (b - c)^2,$ where $a,$ $b,$ and $c$ are the sides of triangle $ABC,$ as usual.  Compute $\tan A.$
extra_info.split: test