Evaluate each expression without using a calculator. (a) \( \tan \left(\arccos \left(\frac{1}{2}\right)\right) \) (b) \( \cos \left(\arcsin \left(\frac{24}{25}\right)\right) \)

To evaluate \( \tan \left(\arccos \left(\frac{1}{2}\right)\right) \), we first find the angle whose cosine is \( \frac{1}{2} \). This angle is \( \frac{\pi}{3} \) or \( 60^\circ \). The tangent of this angle is given by: \[ \tan\left(\frac{\pi}{3}\right) = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \] Therefore, \[ \tan \left(\arccos \left(\frac{1}{2}\right)\right) = \sqrt{3} \] Next, to evaluate \( \cos \left(\arcsin \left(\frac{24}{25}\right)\right) \), we use the relationship between sine and cosine. We know that if \( x = \arcsin \left(\frac{24}{25}\right) \), then \( \sin(x) = \frac{24}{25} \). To find \( \cos(x) \), we can apply the Pythagorean identity: \[ \cos^2(x) + \sin^2(x) = 1 \] Substituting \( \sin(x) \): \[ \cos^2(x) + \left(\frac{24}{25}\right)^2 = 1 \] \[ \cos^2(x) + \frac{576}{625} = 1 \] \[ \cos^2(x) = 1 - \frac{576}{625} = \frac{625 - 576}{625} = \frac{49}{625} \] Taking the square root gives us \( \cos(x) = \frac{7}{25} \) (since cosine is positive for angles in the range of arcsin). Thus, \[ \cos \left(\arcsin \left(\frac{24}{25}\right)\right) = \frac{7}{25} \] In summary, the expressions evaluate to: (a) \( \tan \left(\arccos \left(\frac{1}{2}\right)\right) = \sqrt{3} \) (b) \( \cos \left(\arcsin \left(\frac{24}{25}\right)\right) = \frac{7}{25} \)