Please help me answer the last part of this question below, the explain if the two graphs are the same, THANKS so much!!

show that the solution of:

dy/dx = -((1+y^2)/(1+x^2)) y(0) = -1
is the inverse tan (x) + inverse tangent (y) = -pi/4 AND show that the above can be written :
y= ((x+1)/(x-1))
And are the graphs of:
inverse tan(x) + inverse tan (y) = -pi/4 AND y=((x+1)/(x-1)) from above the same?

Explain.

Question

Please help me answer the last part of this question below, the explain if the two graphs are the same,  THANKS so much!!

show that the solution of: 

dy/dx = -((1+y^2)/(1+x^2)) y(0) = -1 
is the inverse tan (x) + inverse tangent (y) = -pi/4 AND show that the above can be written : 
y= ((x+1)/(x-1)) 
And are the graphs of: 
inverse tan(x) + inverse tan (y) = -pi/4 AND y=((x+1)/(x-1)) from above the same? 

Explain.

dumbcow · Answer

use identity of angle addition with tangent

$$\tan (\tan^{-1} x + \tan^{-1} y) = \tan(-\pi/4)$$

dumbcow · Answer

$$\tan (a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$
$$\rightarrow \frac{x+y}{1-xy} = -1$$

dumbcow · Answer

solving for y will give
$$y = \frac{x+1}{x-1}$$

and yes i believe both graphs would be the same

anonymous · Answer

Do you know why they are the same though?  I dont know how to explain it