观察下列恒等式：∵tan2α-1tanα=-2(1-tan2α)2tanα，∴tanα-1tanα=-2tan2α①∴tan2α-1tan2α=-2tan4α②∴tan4α-1tan4α=-2tan8α③由此可知：tanπ32+2tanπ16+4tanπ8-1tanπ32=．

Question

观察下列恒等式：
∵(tan2α-1)/(tanα)=-(2(1-tan2α))/(2tanα)，
∴tanα-(1)/(tanα)=-(2)/(tan2α)①
∴tan2α-(1)/(tan2α)=-(2)/(tan4α)②
∴tan4α-(1)/(tan4α)=-(2)/(tan8α)③
由此可知：tan(π)/(32)+2tan(π)/(16)+4tan(π)/(8)-(1)/(tan(π)/(32))=______．

Answer 1

∵tanα-1tanα=-2tan2α①∴tan2α-1tan2α=-2tan4α②∴tan4α-1tan4α=-2tan8α③所以tanπ32+2tanπ16+4tanπ8-1tanπ32=tanπ32-1tanπ32+2tanπ16+4tanπ8=-2tanπ16+2tanπ16+4tanπ8=-4tanπ8+4tanπ8=-4×2...