Check the identity: Let x=√3, then tan-1(x)=π/3, cot-1(x)=π/6. sec(π/3)=2, csc(π/6)=2.
So sec2(π/3)+csc2(π/6)=4+4=8; 2(1+x2)=2×4=8. Identity seems OK.
Let y=tan-1(x), so x=tan(y), and z=cot-1(x), so x=cot(z):
sec2(y)+csc2(z)=(1+tan2(y))+(1+cot2(z))=1+x2+1+x2=2(1+x2) QED.
(Note also that x=tan(y), so 1/x=cot(y)=1/cot(z)=tan(z), so cot(y)=tan(z), y+z=π/2.)