use calculus to show cos^-1 (1-x^2)/(1+x^2)=2tan^-1 x
You need to form the function `y =cos^-1 (1-x^2)/(1+x^2)-2tan^-1 (x).`
Notice that the function is the constant 0, hence, differentiating the function, you should also get zero.
You need to differentiate the function with respect to x such that:
`(((1-x^2)/(1+x^2))')/(sqrt(1 - ((1-x^2)/(1+x^2))^2)) - 2/(1 + x^2) = 0`
`((-2x(1+x^2) - 2x(1-x^2))/((1+x^2)^2))/((sqrt((1 + x^2 - 1 + x^2)(1+x^2+1-x^2)))/(1+x^2)) - 2/(1 + x^2) = 0`
`-4x/((1+x^2)sqrt(4x^2)) - 2/(1 + x^2) = 0`
`4x/(2x(1+x^2))- 2/(1 + x^2) = 0`
`2/(1+x^2) - 2/(1 + x^2) = 0`
`0 = 0`
Hence, differentiating the function yields 0
`=> cos^-1 (1-x^2)/(1+x^2)-2tan^-1 (x) = 0 =gt cos^-1 (1-x^2)/(1+x^2) = 2tan^-1 (x).`
Technically the previous answer isn't quite complete:
If you find that the derivative of y is 0, you have found that y is a constant function. But it isn't necessarily the constant zero function. To finish off the problem, you need to figure out which constant, so substitute some value in for x (say, x=0).
Then y(0) = Cos^-1 (1) -2 Tan^-1 (0) = 0
Now you know that not only is y a constant function, but that the constant is 0.
The problem has already provided the information that the function y is the constant zero, hence, it is no more necessary to prove that the value of constant is zero.