Solve inverse tangents How do I solve the following equation:
$$ \tan^{-1}\frac{x}{10^6}+\tan^{-1}\frac{x}{10^7}=90^{\circ}$$
WA
Step by step solution from wolframalpha is unavailable.
 A: The fact that $\alpha + \beta = \frac \pi 2$, suggest that $\alpha$ and $\beta$ are two angles of right triangle, so $\tan \alpha = 1/\tan \beta$, from which
$$
\tan \left( \arctan \frac {x}{10^7}\right ) = \frac 1{\tan \left( \arctan \frac {x}{10^6}\right )} \implies \frac {x}{10^7} = \frac {10^6}x \implies x = 10^6 \sqrt{10}
$$
A: To solve this equation, what you'd first think to do is apply $\tan$ to both sides, to get
$$\tan \left(\tan^{-1} \dfrac{x}{10^6} + \tan^{-1} \dfrac{x}{10^7}\right) = \tan(90^{\circ})$$
but this doesn't work since $\tan(90^{\circ})$ is undefined.
But look at the left-hand side and compare it to the identity
$$\tan(A+B) = \dfrac{\tan A + \tan B}{1-\tan A\tan B}$$
Here we'd set $A = \tan^{-1}\dfrac{x}{10^6}$ and $B = \tan^{-1}\dfrac{x}{10^7}$.
For this to be undefined in this case, the denominator must be equal to zero.
So
$$\tan^{-1} \dfrac{x}{10^6} + \tan^{-1} \dfrac{x}{10^7} = 90^{\circ} \quad \Leftrightarrow \quad 1- \left(\dfrac{x}{10^6}\right) \left( \dfrac{x}{10^7} \right) = 0$$
I leave the rest to you.
A: Hints: first apply
1)
$$ \arctan 1/x = \pi/2 - \arctan x, \quad x> 0 $$
then
2) apply $\tan x $ to both sides of the equation.
