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`        The sides a,b,c of triangle ABC are in A.P. if cosα = a/b+c , cosβ = b/c+a , cos gama =c/a+b then tan^2(α/2)+tan^2(gama/2)is equal to `
one year ago

```							we know that tan^2(x/2)= (1 – cosx)/(1+cosx)now, tan^2(α/2)= (1 – cosα)/(1+cosα)= (b+c-a)/(b+c+a)and tan^2(gama/2)= (1 – cosgama)/(1+cosgama)= (a+b-c)/(a+b+c)addingtan^2(α/2)+tan^2(gama/2)= 2b/(a+b+c)but a+c=2bso tan^2(α/2)+tan^2(gama/2)= 2b/3b= 2/3
```
one year ago
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