If tanA= 1 √x(x^2+x+1), tanB= √x √x^2+x+1 and tanC=(x^-3+x^-2+x^-1)^ 1 2,0<A,B,C< π 2, then A+B is equal to:

Given, $\mathrm{tan}A=\frac{1}{\sqrt{x({x}^{2}+x+1)}},\mathrm{tan}B=\frac{\sqrt{x}}{\sqrt{{x}^{2}+x+1}}$tanA=x(x2+x+1)​1​,tanB=x2+x+1​x​​ and $\mathrm{tan}C={({x}^{-3}+{x}^{-2}+{x}^{-1})}^{\frac{1}{2}}=\frac{\sqrt{1+x+{x}^{2}}}{x\sqrt{x}}$tanC=(x−3+x−2+x−1)21​=xx​1+x+x2​​We know that,

$$\mathrm{tan}(A+B)=\frac{\mathrm{tan}A+\mathrm{tan}B}{1-\mathrm{tan}A\mathrm{tan}B}$$tan(A+B)=1−tanAtanBtanA+tanB​

$$\Rightarrow \mathrm{tan}(A+B)=\frac{\frac{1}{\sqrt{x({x}^{2}+x+1)}}+\frac{\sqrt{x}}{\sqrt{{x}^{2}+x+1}}}{1-\frac{1}{{x}^{2}+x+1}}$$⇒tan(A+B)=1−x2+x+11​x(x2+x+1)​1​+x2+x+1​x​​​

$$\Rightarrow \mathrm{tan}(A+B)=\frac{(1+x)(\sqrt{{x}^{2}+x+1})}{({x}^{2}+x)(\sqrt{x})}$$⇒tan(A+B)=(x2+x)(x​)(1+x)(x2+x+1​)​

$$\Rightarrow \mathrm{tan}(A+B)=\frac{(1+x)(\sqrt{{x}^{2}+x+1})}{({x}^{2}+x)(\sqrt{x})}$$⇒tan(A+B)=(x2+x)(x​)(1+x)(x2+x+1​)​

$$\Rightarrow \mathrm{tan}(A+B)=\frac{\sqrt{{x}^{2}+x+1}}{x\sqrt{x}}$$⇒tan(A+B)=xx​x2+x+1​​

$$\Rightarrow \mathrm{tan}(A+B)=\mathrm{tan}C$$⇒tan(A+B)=tanC

$$\Rightarrow A+B=C$$⇒A+B=C