# Tangent line problem with implicit differentiation

Given:

$[\tan^{-1}(x)]^2+[\cot^{-1}(y)]^2=1$

Find the tangent line equation to the graph at the point $(1,0)$ by implicit differentiation

I found the derivative:

$\dfrac{dy}{dx}=\dfrac{4\tan^{-1}(x)\cdot \cot^{-1}(y)}{(y^2+1)(x^2+1)}$

I may have done my derivative wrong, but my main concern is at some point $0$ will be plugged into $\cot$ inverse, resulting in division by zero.

I need help with this scenario.

• Why do you say that evaluating the inverse cotangent function at 0 will result in a division by zero? $$\cot^{-1}(0) = \frac{\pi}{2}$$ Feb 27 '16 at 4:39
• You know I clearly mixed things up, thanks. Feb 27 '16 at 4:52
• $(\tan^{-1} 1)^{2}+(\cot^{-1} 0)^{2}=\frac{5\pi^{2}}{16} \neq 1$ Feb 27 '16 at 9:25

$\displaystyle \frac{2\tan^{-1} x}{1+x^{2}} \, dx-\frac{2\cot^{-1} y}{1+y^{2}} \, dy=0$
$\displaystyle \frac{dy}{dx}=\frac{(1+y^{2})\tan^{-1} x}{(1+x^{2})\cot^{-1} y}$
Note that $(1,0)$ doesn't lie on the graph.
The graph can be parametrized as $(x,y)=(\tan \cos t,\cot \sin t)$