Draw the graph of `y=cos^(-1).(1-x^(2))/(1+x^(2))`.

Explanation

We have `y=f(x)=cos^(2)((1+x^(2))/(1+x^(2)))`
Let `x=tantheta, theta in (-pi//2, pi//2)`
`rArr" "theta=tan^(-1)x`
Now `cos^(-1)((1-x^(2))/(1+x^(2)))=cos^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))`
`=cos^(-1)(cos2theta)`
`=cos^(-1)(cos2theta)," where "alpha in (-pi, pi)`
Now consider the graph of `y=cos^(-1)(cos alpha)," where "alpha in (-pi, pi)`

From the graph.
`cos^(-1)((1-x^(2))/(1+x^(2)))=cos^(-1)(cosalpha)`
`={{:(-alpha,-piltalphalt0),(alpha,0lealphaltpi):}`
`={{:(-2tan^(-1)x,-pilt2tan^(-1)xlt0),(2tan^(-1)x, 0le2tan^(-1)xltpi):}`
`={{:(-2tan^(-1)x,-pilt2tan^(-1)xlt0),(2tan^(-1)x, 0le2tan^(-1)xltpi//2):}`
`={{:(-2tan^(-1)x,xlt0),(2tan^(-1)x,xge0):}`
`f'(x)={{:(-2/(1+x^(2)),xlt0),(2/(1+x^(2)),xgt0):}`
So f(x) is non-differentiable at x = 0
`underset(xtooo)(lim)(-2tan^(-1)x)=pi,f(0)=2tan^(-1)(0)=0`
Thus, `cos^(-1)((1-x^(2))/(1+x^(2)))" decreases form "pi" to "0" as increases from"-oo" to 0".`
`underset(xtooo)(lim)(2tan^(-1)x)=pi`
Thus, `cos^(-1)((1-x^(2))/(1+x^(2)))" increases form 0 to "pi" as x increses from 0 to"oo`.
From this information, we can draw the graph of `y=cos^(-1)((1-x^(2))/(1+x^(2)))` as shown in the following figure.