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`        If tana=(1+2^-x)^-1,.             tanb=(1+2^(x+1))^-1then.  A+b=`
one year ago

```							answer of the question is 45hint=find tan(a+b) put the value tana ‘tanband slovegot tan[a+b]=1[a+b]=1
```
one year ago
```							Tan(a+b)=tana + tanb/1-tanatanb =(1+2^-x)^-1 + (1+2^x+1)^-1/1-(1+2^-x)^-1.(1+2^x+1)^-1 =1+2^x+1+2^-x-1 / 1-(1+2^x)(1+2^-x-1)=[(2^x+2 + 2^2x+1 + 1)/2^x+1]/[(2^2x+1 + 2^x+1 +1)/2^x+1]=1=45°=π/4
```
one year ago
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