Find the exact value of the expression. \[ \cos \left[\tan ^{-1} \frac{15}{8}-\cos ^{-1} \frac{7}{25}\right] \] \( \cos \left[\tan ^{-1} \frac{15}{8}-\cos ^{-1} \frac{7}{25}\right]=\square \) (Type an integer or a simplified fraction.)

To find the exact value of the expression \[ \cos \left[\tan ^{-1} \frac{15}{8}-\cos ^{-1} \frac{7}{25}\right], \] we can use the cosine subtraction formula: \[ \cos(A - B) = \cos A \cos B + \sin A \sin B. \] Let: - \( A = \tan^{-1} \frac{15}{8} \) - \( B = \cos^{-1} \frac{7}{25} \) ### Step 1: Calculate \( \cos A \) and \( \sin A \) For \( A = \tan^{-1} \frac{15}{8} \), we can interpret this as a right triangle where: - Opposite side = 15 - Adjacent side = 8 The hypotenuse can be calculated using the Pythagorean theorem: \[ \text{Hypotenuse} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17. \] So, we can find \( \cos A \) and \( \sin A \): \[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{17}, \] \[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{15}{17}. \] ### Step 2: Calculate \( \cos B \) and \( \sin B \) For \( B = \cos^{-1} \frac{7}{25} \), we know: \[ \cos B = \frac{7}{25}. \] Using the Pythagorean identity: \[ \sin^2 B + \cos^2 B = 1 \] \[ \sin^2 B + \left(\frac{7}{25}\right)^2 = 1 \] \[ \sin^2 B + \frac{49}{625} = 1 \] \[ \sin^2 B = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}. \] Thus, \[ \sin B = \sqrt{\frac{576}{625}} = \frac{24}{25}. \] ### Step 3: Combine using the cosine subtraction formula Now we can substitute \( \sin A \), \( \sin B \), \( \cos A \), and \( \cos B \) into the cosine subtraction formula: \[ \cos \left[\tan^{-1} \frac{15}{8} - \cos^{-1} \frac{7}{25}\right] = \cos A \cos B + \sin A \sin B \] \[ = \left(\frac{8}{17} \cdot \frac{7}{25}\right) + \left(\frac{15}{17} \cdot \frac{24}{25}\right). \] Calculating each term: \[ = \frac{56}{425} + \frac{360}{425} = \frac{56 + 360}{425} = \frac{416}{425}. \] Thus, the exact value of the expression is \[ \boxed{\frac{416}{425}}. \]