If operatorname{Sinh}^{-1} x = operatorname{C | vedclass

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Question

(A) Given $\operatorname{Sinh}^{-1} x = \log(1+\sqrt{2})$.Using the definition $\operatorname{Sinh}^{-1} x = \log(x + \sqrt{x^2+1})$, we have $x + \sq

Vedclass · Accepted Answer

$67 \frac{1}{2}^{\circ}$

Vedclass · Answer

(A) Given $\operatorname{Sinh}^{-1} x = \log(1+\sqrt{2})$.Using the definition $\operatorname{Sinh}^{-1} x = \log(x + \sqrt{x^2+1})$, we have $x + \sqrt{x^2+1} = 1+\sqrt{2}$.Comparing the terms, we get $x = 1$.Given $\operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$.Using the definition $\operatorname{Cosh}^{-1} y = \log(y + \sqrt{y^2-1})$, we have $y + \sqrt{y^2-1} = 1+\sqrt{2}$.This implies $y = \sqrt{2}$.Now, we need to find $\operatorname{Tan}^{-1}(x+y) = \operatorname{Tan}^{-1}(1+\sqrt{2})$.We know that $	an(67.5^{\circ}) = 	an(\frac{135^{\circ}}{2}) = \frac{1-\cos(135^{\circ})}{\sin(135^{\circ})} = \frac{1 - (-1/\sqrt{2})}{1/\sqrt{2}} = \sqrt{2}+1$.Therefore, $\operatorname{Tan}^{-1}(1+\sqrt{2}) = 67.5^{\circ} = 67 \frac{1}{2}^{\circ}$.

If $\operatorname{Sinh}^{-1} x = \operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$, then $\operatorname{Tan}^{-1}(x+y) = $

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