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

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Aditya Gupta · Answer

we know that tan^2(x/2)= (1 – cosx)/(1+cosx) now, tan^2(α/2)= (1 – cos α )/(1+cos α )= (b+c-a)/(b+c+a) and tan^2(gama/2)= (1 – cos gama )/(1+cos gama )= (a+b-c)/(a+b+c) adding tan^2(α/2)+tan^2(gama/2)= 2b/( a+b+c) but a+c=2b so tan^2(α/2)+tan^2(gama/2)= 2b/3b= 2/3

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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

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

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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

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

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