Find the values of the remaining trigonometric functions when cos theta \( =9 / 41 \) and the terminal arm of the angle is in quad IV

To find the values of the remaining trigonometric functions given that \(\cos \theta = \frac{9}{41}\) and that the terminal arm of the angle is in the fourth quadrant, we can use the Pythagorean identity and the definitions of the trigonometric functions. 1. **Find \(\sin \theta\)**: We can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \(\cos \theta\): \[ \sin^2 \theta + \left(\frac{9}{41}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{81}{1681} = 1 \] \[ \sin^2 \theta = 1 - \frac{81}{1681} \] \[ \sin^2 \theta = \frac{1681 - 81}{1681} = \frac{1600}{1681} \] \[ \sin \theta = \pm \sqrt{\frac{1600}{1681}} = \pm \frac{40}{41} \] Since \(\theta\) is in the fourth quadrant, where sine is negative: \[ \sin \theta = -\frac{40}{41} \] 2. **Find \(\tan \theta\)**: The tangent function is defined as: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Substituting the values we found: \[ \tan \theta = \frac{-\frac{40}{41}}{\frac{9}{41}} = -\frac{40}{9} \] 3. **Find \(\csc \theta\)**: The cosecant function is the reciprocal of sine: \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{40}{41}} = -\frac{41}{40} \] 4. **Find \(\sec \theta\)**: The secant function is the reciprocal of cosine: \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{9}{41}} = \frac{41}{9} \] 5. **Find \(\cot \theta\)**: The cotangent function is the reciprocal of tangent: \[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{40}{9}} = -\frac{9}{40} \] Now we have all the trigonometric functions: - \(\sin \theta = -\frac{40}{41}\) - \(\cos \theta = \frac{9}{41}\) - \(\tan \theta = -\frac{40}{9}\) - \(\csc \theta = -\frac{41}{40}\) - \(\sec \theta = \frac{41}{9}\) - \(\cot \theta = -\frac{9}{40}\)