It's based on how they are defined and the nature of the graphs of sine, cosine, and tangent (I'll assume you are familiar with their graphs in what follows).
For example, #x=f^(-1}(y)=sin^{-1}(y)# is defined to be the inverse function of #y=f(x)=sin(x)# for #-pi/2\leq x\leq pi/2#. Since #y=f(x)=sin(x)# is continuous and #y->1# as #x->\frac{pi}{2}^{-}# (the minus sign to the right of the number indicates the number is being approached "from the left"), it follows that #x=f^{-1}(y)=sin^{-1}(y)->pi/2# as #y->1^{-}#. Similarly, #x=f^{-1}(y)=sin^{-1}(y)->-\pi/2# as #y->-1^{+}#.
Since #x=g^{-1}(y)=cos^{-1}(y)# is defined to be the inverse function of #y=g(x)=cos(x)# for #0\leq x\leq pi# and both functions are continuous, it follows that #x=g^{-1}(y)=cos^{-1}(y)->0# as #y->1^{-}# and #x=g^{-1}(y)=cos^{-1}(y)->pi# as #y->-1^{+}#.
Since #x=h^{-1}(y)=tan^{-1}(y)# is defined to be the inverse function of #y=h(x)=tan(x)# for #-pi/2 < x < pi/2# and both functions are continuous, it follows that #x=h^{-1}(y)=tan^{-1}(y)->pi/2# as #y->+\infty# and #x=h^{-1}(y)=tan^{-1}(y)->-pi/2# as #y->-\infty#.
