已知(1-tana)/(2+tana)=1
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已知(1-tana)/(2+tana)=1
求证tan2a=-4tan(a+派/4)
求证tan2a=-4tan(a+派/4)
1-tanA/2+tanA=1
1-tanA=2+tanA
tanA=-1/2;
tan2A
=2tanA/(1-tan^2 A)
=2*(-1/2)/[1-(-1/2)^2]
=-1/[1-1/4]
=-1/(3/4)
=-4/3
-4tan(A+π/4)
=-4*[(tanA+tanπ/4)/(1-tanAtanπ/4)]
=-4*[(tanA+1)/(1-tanA)]
=-4*[(-1/2+1)/(1+1/2)]
=-4*[(1/2)/(3/2)]
=-4*1/3
=-4/3
所以tan2A=-4tan(A+π/4)
1-tanA=2+tanA
tanA=-1/2;
tan2A
=2tanA/(1-tan^2 A)
=2*(-1/2)/[1-(-1/2)^2]
=-1/[1-1/4]
=-1/(3/4)
=-4/3
-4tan(A+π/4)
=-4*[(tanA+tanπ/4)/(1-tanAtanπ/4)]
=-4*[(tanA+1)/(1-tanA)]
=-4*[(-1/2+1)/(1+1/2)]
=-4*[(1/2)/(3/2)]
=-4*1/3
=-4/3
所以tan2A=-4tan(A+π/4)
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求证tan(A/2)-{1/(tanA/2)}=-2/tanA
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