JEE Main 20241 Feb, Morning shiftMathematicsHardTrigonometry118× since 2002If tanA=1x(x2+x+1),tanB=xx2+x+1\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}tanA=x(x2+x+1)1,tanB=x2+x+1x and tanC=(x−3+x−2+x−1)1/2,0<A,B,C<π2\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}tanC=(x−3+x−2+x−1)1/2,0<A,B,C<2π, then A+B\mathrm{A}+\mathrm{B}A+B is equal to :AOption A. C\mathrm{C}CBOption B. π−C\pi-Cπ−CCOption C. 2π−C2 \pi-C2π−CDOption D. π2−C\frac{\pi}{2}-\mathrm{C}2π−CCheckSkip · show solution
