1. **Problem Statement:** Evaluate the expression $$\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arctan(1) - \arcsec(\sqrt{2})$$ and verify if it equals $$\frac{\pi}{4}$$. 2. **Recall the definitions and principal values:** - $$\arcsin(x)$$ is the inverse sine function with principal values in $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. - $$\arctan(x)$$ is the inverse tangent function with principal values in $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. - $$\arcsec(x)$$ is the inverse secant function, principal values usually taken in $$[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}$$. 3. **Evaluate each term:** - $$\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$$ because $$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$. - $$\arctan(1) = \frac{\pi}{4}$$ because $$\tan\left(\frac{\pi}{4}\right) = 1$$. - $$\arcsec(\sqrt{2})$$ means find $$\theta$$ such that $$\sec(\theta) = \sqrt{2}$$ and $$\theta \in [0, \pi] \setminus \left\{\frac{\pi}{2}\right\}$$. Since $$\sec(\theta) = \frac{1}{\cos(\theta)}$$, we have $$\cos(\theta) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$. The angle $$\theta$$ in $$[0, \pi]$$ with $$\cos(\theta) = \frac{\sqrt{2}}{2}$$ is $$\frac{\pi}{4}$$. 4. **Substitute values:** $$\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arctan(1) - \arcsec(\sqrt{2}) = \frac{\pi}{3} + \frac{\pi}{4} - \frac{\pi}{4} = \frac{\pi}{3}$$ 5. **Compare with $$\frac{\pi}{4}$$:** The expression equals $$\frac{\pi}{3}$$, which is not equal to $$\frac{\pi}{4}$$. **Final answer:** The assertion that the expression equals $$\frac{\pi}{4}$$ is **false**; the correct value is $$\frac{\pi}{3}$$.