Find the exact value of the expression. \[ \sin \left[\sin ^{-1} \frac{3}{5}+\cos ^{-1}\left(-\frac{7}{25}\right)\right] \] \( \sin \left[\sin ^{-1} \frac{3}{5}+\cos ^{-1}\left(-\frac{7}{25}\right)\right]=\square \)

To solve the expression \( \sin \left[\sin ^{-1} \frac{3}{5}+\cos ^{-1}\left(-\frac{7}{25}\right)\right] \), we can start by letting: \[ x = \sin^{-1}\left(\frac{3}{5}\right) \] This gives us \( \sin x = \frac{3}{5} \). Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \), we can find \( \cos x \): \[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] Thus, \( \cos x = \frac{4}{5} \) (since the angle is in the range of \( \sin^{-1} \), we take the positive root). Next, we evaluate \( \cos^{-1}\left(-\frac{7}{25}\right) \). Let: \[ y = \cos^{-1}\left(-\frac{7}{25}\right) \] This gives us \( \cos y = -\frac{7}{25} \). Again, using the Pythagorean identity, we can find \( \sin y \): \[ \sin^2 y = 1 - \cos^2 y = 1 - \left(-\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625} \] Thus, \( \sin y = \frac{24}{25} \) (since \( y \) lies in the second quadrant where sin is positive). Now, we can find \( \sin(x + y) \) using the sine addition formula: \[ \sin(x + y) = \sin x \cos y + \cos x \sin y \] Substituting the values we found: \[ \sin(x + y) = \left(\frac{3}{5}\right)\left(-\frac{7}{25}\right) + \left(\frac{4}{5}\right)\left(\frac{24}{25}\right) \] Calculating each term: First term: \[ \frac{3}{5} \cdot -\frac{7}{25} = -\frac{21}{125} \] Second term: \[ \frac{4}{5} \cdot \frac{24}{25} = \frac{96}{125} \] Now, summing these: \[ \sin(x + y) = -\frac{21}{125} + \frac{96}{125} = \frac{75}{125} = \frac{3}{5} \] Therefore, the exact value of the expression is: \[ \boxed{\frac{3}{5}} \]