We are given that: \[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}. \] The range of the inverse sine function is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), so the maximum possible value of each of the individual terms \( \sin^{-1} x \), \( \sin^{-1} y \), and \( \sin^{-1} z \) is \( \frac{\pi}{2} \).
Step 1: Since the sum of three inverse sines equals \( \frac{3\pi}{2} \), we conclude that each of the terms must be equal to \( \frac{\pi}{2} \). Thus: \[ \sin^{-1} x = \frac{\pi}{2}, \quad \sin^{-1} y = \frac{\pi}{2}, \quad \sin^{-1} z = \frac{\pi}{2}. \] Step 2: Taking the sine of both sides of these equations, we find: \[ x = \sin \frac{\pi}{2} = 1, \quad y = \sin \frac{\pi}{2} = 1, \quad z = \sin \frac{\pi}{2} = 1. \] Step 3: Therefore, the sum \( x + y + z \) is: \[ x + y + z = 1 + 1 + 1 = 3. \]
Thus, the correct answer is option (E).
Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \). Then:
For \(1 \leq x<\infty\), let \(f(x) = \sin^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{x}\right)\). Then \(f'(x) =\)
\[ \int_0^{\frac{\pi}{4}} (\tan^3 x + \tan^5 x) \, dx \]