Step 1: Find \( x \) and \( y \):
Using the logarithmic definitions of inverse hyperbolic functions:
1. \( \sinh^{-1}x = \log(x + \sqrt{x^2+1}) \)
Given \( \log(x + \sqrt{x^2+1}) = \log(1+\sqrt{2}) \).
Comparing terms, clearly \( x=1 \).
2. \( \cosh^{-1}y = \log(y + \sqrt{y^2-1}) \)
We need this to equal \( \log(1+\sqrt{2}) = \log(\sqrt{2}+1) \).
Comparing \( y + \sqrt{y^2-1} \) with \( \sqrt{2} + 1 \), we set \( y = \sqrt{2} \).
Check: \( \sqrt{2} + \sqrt{2-1} = \sqrt{2}+1 \). Correct.
Step 2: Evaluate \( \tan^{-1}(x+y) \):
\[ x+y = 1 + \sqrt{2} \]
We need to find angle \( \theta \) such that \( \tan\theta = \sqrt{2}+1 \).
Recall that \( \tan(22.5^\circ) = \sqrt{2}-1 \) and \( \cot(22.5^\circ) = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1 \).
Since \( \cot(22.5^\circ) = \tan(90^\circ - 22.5^\circ) = \tan(67.5^\circ) \).
Thus, \( \theta = 67.5^\circ = 67\frac{1}{2}^\circ \).