tanα=3^α tanβ =3^(-α )且α -β =π/6 则α 等于
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tanα=3^α tanβ =3^(-α )且α -β =π/6 则α 等于
tanα=3^α tanβ =3^(-α )且α -β =π/6 则α 等于
tanα=3^α tanβ =3^(-α )且α -β =π/6 则α 等于
tan(A-B)=(tanA-tanB)/(1+tanAtanB)=√3/3
3^α-3^(-α )/(1+3^α*3^(-α ))=√3/3
3^α-3^(-α )/2=√3/3
令x=3^α x-1/x)=2√3/3
x^2-2√3/3*x-1=0
x=√3 ∴a=1/2
证明:tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
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