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已知sinB=msin(2x+B)且x+B≠kπ+π/2,(k∈z),x≠kπ/2 (k∈z),m≠1.求证:tan(x

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已知sinB=msin(2x+B)且x+B≠kπ+π/2,(k∈z),x≠kπ/2 (k∈z),m≠1.求证:tan(x+B)=(1+m/1-m)t
接上:nx
已知sinB=msin(2x+B)且x+B≠kπ+π/2,(k∈z),x≠kπ/2 (k∈z),m≠1.求证:tan(x
已知sinB=msin(2x+B)且x+B≠kπ+π/2,(k∈z),x≠kπ/2 (k∈z),m≠1.
求证:tan(x+B)=[(1+m)/(1-m)]tanx
证明:∵sinB=msin(2x+B),∴m=sinB/sin(2x+B)
故(1+m)/(1-m)=[1+sinB/sin(2x+B)]/[1-sinB/sin(2x+B)]=[sin(2x+B)+sinB]/[sin(2x+B)-sinB]
=[2sin(x+B)cosx]/[2cos(x+B)sinx]=tan(x+B)/tanx
∴tan(x+B)=[(1+m)/(1-m)]tanx.故证.