Question

tan^-1 y=tan^-1 x+C

Solve for y.

Answer 1

@SithsAndGiggles

Answer 2

No initial values?

Anyway, just take the tangent of both sides:
\[\tan(\tan^{-1}y)=\tan(\tan^{-1}x+C)~~\iff~~y=\tan(\tan^{-1}x+C)\]

Answer 3

Wait a minute. Let me work it out.

Answer 4

This is the full problem:

Solve y'=(1+y^2)/(1+x^2) explicitly. HINT: Use the identity tan(A+B)=(tanA+tanB)/(1-tanAtanB).

Answer 5

Okay, so using the identity gives you
\[\begin{align*}
y&=\tan(\tan^{-1}x+C)\\\\
&=\frac{\tan(\tan^{-1}x)+\tan C}{1-\tan(\tan^{-1}x)\tan C}\\\\
&=\frac{x+\tan C}{1-x\tan C}\\\\

\end{align*}\]

Answer 6

Thank you so much! Sorry the internet runs a bit slow.

Answer 7

You're welcome!