已知a为锐角,tan(a+b)=3,tan(a-b)=2,则a为?

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已知a为锐角,tan(a+b)=3,tan(a-b)=2,则a为?

已知a为锐角,tan(a+b)=3,tan(a-b)=2,则a为?
已知a为锐角,tan(a+b)=3,tan(a-b)=2,则a为?

已知a为锐角,tan(a+b)=3,tan(a-b)=2,则a为?
a、b都是锐角,则:
tan2a=tan[(a+b)+(a-b)]
=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]
=-1
因为:0<2a<180°
则:2a=3π/4
得:a=3π/8

tan2a=[tan(a+b)+tan(a-b)]/[1-tan(a+b)*tan(a-b)]
tan2a=-1
2a=3π/4
a=3π/8

tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=3
tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)=2
tan2α=tan[(α+β)+(α-β)]=[tan(α+β)+tan(α-β)]/[1-tan(α+β)tan(α-β)]=-1
∵0<2α<180°
∴α=3π/8

tan(a+b)=3,
(tana+tanb)/(1-tanatanb)=3
tan(a-b)=2
(tana-tanb)/(1+tanatanb)=2
tan2a=-1
a为锐角,故:2a=3π/4
a=3π/8

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