Hard· Properties of inverse trigonometric functions· +4 / −1If tanA=1x2+x+1\tan A=\frac{1}{\sqrt{x^2+x+1}}tanA=x2+x+11, tanB=xx2+x+1\tan B=\frac{\sqrt x}{\sqrt{x^2+x+1}}tanB=x2+x+1x and tanC=1x(x2+x+1)\tan C=\frac{1}{\sqrt{x(x^2+x+1)}}tanC=x(x2+x+1)1, then A+B=A+B=A+B=Aπ2−C\frac\pi2-C2π−CBπ−C\pi-Cπ−CC0DNoneCheck answerJust show me the answer
