\( \theta_{1}=\theta_{2}=\theta_{2}=2 \) If \( 1+\sin \theta=9 \cos \theta \) and \( 0^{\circ}<\theta<90^{\circ} \), then analyse and assess the value of \( 5(1-\cos \theta) \). \( 1+\sin \theta=9 \cos \theta \)

We are given the equation \[ 1 + \sin \theta = 9 \cos \theta \] with \(0^\circ < \theta < 90^\circ\) and need to find the value of \[ 5(1 - \cos \theta). \] **Step 1. Solve for \(\sin \theta\) and \(\cos \theta\).** Starting with \[ 1 + \sin \theta = 9 \cos \theta, \] we can rearrange this as \[ \sin \theta = 9 \cos \theta - 1. \] **Step 2. Square both sides and use the Pythagorean identity.** Square the equation: \[ (1 + \sin \theta)^2 = (9 \cos \theta)^2. \] Expanding and simplifying, we have: \[ 1 + 2 \sin \theta + \sin^2 \theta = 81 \cos^2 \theta. \] Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\sin^2 \theta = 1 - \cos^2 \theta\): \[ 1 + 2 \sin \theta + (1 - \cos^2 \theta) = 81 \cos^2 \theta. \] This simplifies to: \[ 2 + 2 \sin \theta - \cos^2 \theta = 81 \cos^2 \theta. \] Bring the \(\cos^2 \theta\) term to the right: \[ 2 + 2 \sin \theta = 82 \cos^2 \theta. \] Although we could proceed by expressing \(\cos^2 \theta\) in terms of \(\sin \theta\), a more straightforward method is to find \(\sin \theta\) directly by another approach. **Step 3. Solve by forming a quadratic in \(\sin \theta\).** Express the original equation \(1 + \sin \theta = 9 \cos \theta\) in a different way by isolating \(\cos \theta\): \[ \cos \theta = \frac{1+\sin \theta}{9}. \] Now substitute this in the identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \] Thus, \[ \sin^2 \theta + \left(\frac{1+\sin \theta}{9}\right)^2 = 1. \] Multiply through by \(81\) (which is \(9^2\)) to eliminate the denominator: \[ 81 \sin^2 \theta + (1+\sin \theta)^2 = 81. \] Expand \((1+\sin \theta)^2\): \[ 81 \sin^2 \theta + 1 + 2 \sin \theta + \sin^2 \theta = 81. \] Combine like terms: \[ 82 \sin^2 \theta + 2 \sin \theta + 1 - 81 = 0, \] \[ 82 \sin^2 \theta + 2 \sin \theta - 80 = 0. \] Divide the entire equation by \(2\): \[ 41 \sin^2 \theta + \sin \theta - 40 = 0. \] Let \(t = \sin \theta\). The quadratic is \[ 41t^2 + t - 40 = 0. \] The discriminant is: \[ \Delta = 1^2 - 4(41)(-40) = 1 + 6560 = 6561. \] Since \(\sqrt{6561} = 81\), the solutions for \(t\) are: \[ t = \frac{-1 \pm 81}{2 \cdot 41}. \] This gives: \[ t = \frac{80}{82} = \frac{40}{41} \quad \text{or} \quad t = \frac{-82}{82} = -1. \] Since \(0^\circ < \theta < 90^\circ\), we take the positive solution: \[ \sin \theta = \frac{40}{41}. \] **Step 4. Find \(\cos \theta\).** Substitute the value of \(\sin \theta\) back into the equation: \[ 1 + \sin \theta = 9 \cos \theta. \] Thus, \[ 1+\frac{40}{41} = 9 \cos \theta, \] \[ \frac{41+40}{41} = 9\cos \theta, \] \[ \frac{81}{41} = 9\cos \theta. \] So, \[ \cos \theta = \frac{81}{41 \cdot 9} = \frac{9}{41}. \] **Step 5. Compute \(5(1-\cos \theta)\).** Now, substitute \(\cos \theta = \frac{9}{41}\) into the expression: \[ 5(1-\cos \theta) = 5\left(1-\frac{9}{41}\right). \] Simplify inside the parentheses: \[ 1-\frac{9}{41} = \frac{41-9}{41} = \frac{32}{41}. \] Thus, \[ 5(1-\cos \theta) = 5 \cdot \frac{32}{41} = \frac{160}{41}. \] \[ \boxed{\frac{160}{41}} \]