判断y=lg[(tanx+1)/(tanx-1)]的奇偶性

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判断y=lg[(tanx+1)/(tanx-1)]的奇偶性判断y=lg[(tanx+1)/(tanx-1)]的奇偶性判断y=lg[(tanx+1)/(tanx-1)]的奇偶性令f(x)=lg[(tan

判断y=lg[(tanx+1)/(tanx-1)]的奇偶性
判断y=lg[(tanx+1)/(tanx-1)]的奇偶性

判断y=lg[(tanx+1)/(tanx-1)]的奇偶性
令f(x)=lg[(tanx+1)/(tanx-1)]
f(-x)=lg[(tan(-x)+1)/(tan(-x)-1)]
=lg[(tanx-1)/(tanx+1)]
=-lg[tanx+1)/(tanx-1)]
所以是奇函数