Questions: Find the exact value. Simplify your answer completely. Write any fractional answers as a single fraction in simplest form. Rationalize your denominator, if necessary. tan(arccos(1/5)-arcsin(3/5))=

Solution Steps

To find \(\tan \left(\arccos \frac{1}{5} - \arcsin \frac{3}{5}\right)\), we can use the tangent subtraction formula: \(\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}\). First, we need to find \(\tan(\arccos \frac{1}{5})\) and \(\tan(\arcsin \frac{3}{5})\). For \(\tan(\arccos \frac{1}{5})\), use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Similarly, find \(\tan(\arcsin \frac{3}{5})\) using the same identities. Then, substitute these values into the tangent subtraction formula.

Let \( a = \arccos \frac{1}{5} \). We know that: \[ \cos a = \frac{1}{5} \] Using the Pythagorean identity, we find \(\sin a\): \[ \sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \] Thus, we can calculate \(\tan a\): \[ \tan a = \frac{\sin a}{\cos a} = \frac{\frac{2\sqrt{6}}{5}}{\frac{1}{5}} = 2\sqrt{6} \]

Let \( b = \arcsin \frac{3}{5} \). We know that: \[ \sin b = \frac{3}{5} \] Using the Pythagorean identity, we find \(\cos b\): \[ \cos b = \sqrt{1 - \sin^2 b} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \] Thus, we can calculate \(\tan b\): \[ \tan b = \frac{\sin b}{\cos b} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

Now we can use the tangent subtraction formula: \[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \] Substituting the values we found: \[ \tan(a - b) = \frac{2\sqrt{6} - \frac{3}{4}}{1 + 2\sqrt{6} \cdot \frac{3}{4}} \] Calculating the numerator: \[ 2\sqrt{6} - \frac{3}{4} = \frac{8\sqrt{6}}{4} - \frac{3}{4} = \frac{8\sqrt{6} - 3}{4} \] Calculating the denominator: \[ 1 + 2\sqrt{6} \cdot \frac{3}{4} = 1 + \frac{3\sqrt{6}}{2} = \frac{2 + 3\sqrt{6}}{2} \] Thus, we have: \[ \tan(a - b) = \frac{\frac{8\sqrt{6} - 3}{4}}{\frac{2 + 3\sqrt{6}}{2}} = \frac{8\sqrt{6} - 3}{4} \cdot \frac{2}{2 + 3\sqrt{6}} = \frac{2(8\sqrt{6} - 3)}{4(2 + 3\sqrt{6})} = \frac{8\sqrt{6} - 3}{2(2 + 3\sqrt{6})} \]

Final Answer