tan(4/π+α)=3 求tanα?

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tan(4/π+α)=3求tanα?tan(4/π+α)=3求tanα?tan(4/π+α)=3求tanα?tana=tan[(4/π+α)-4/π]=(tan(4/π+α)-tan4/π)/(1+t

tan(4/π+α)=3 求tanα?
tan(4/π+α)=3 求tanα?

tan(4/π+α)=3 求tanα?
tana=tan[(4/π+α)-4/π]
=(tan(4/π+α)-tan4/π)/(1+tan(4/π+α)*tan4/π)
=(3-1)/(1+3)=1/2

tan(π/4+α)=3
(tanπ/4+tanα)/(1-tanπ/4tanα)=3
(1+tanα)/(1-tanα)=3
1+tanα=3(1-tanα)
1+tanα=3-3tanα
4tanα=2
tanα=1/2

tan(4/π+α)=(tan4/π+tanα)/(1-tan4/πtanα)=(1+tanα)/(1-tanα)=3
解得tanα=1/2

tan(4/π+α)=(1+tanα)/(1-tanα)=3 tanα=1/2