Prove that: \( \frac{1}{\tan } a(\sin a \tan a+\cos a)=\frac{1}{\sin a} \) If \( (4 \theta-8) \sin 30^{\circ}=\left(\theta^{3}-8\right) \) and \( \left(Q^{2}+2 \theta+4\right)=2 \), determine value ofs \( \tan 240^{\circ} \), withoud the use of a calculator Determine the general Solution of: a) \( 2 \cos 2 \theta=-0,44 \) b) \( 2 \cos ^{2} \theta-\cos Q=0 \)

Let's break down the problem step by step. ### Part 1: Prove that \( \frac{1}{\tan a}(\sin a \tan a + \cos a) = \frac{1}{\sin a} \) 1. **Start with the left-hand side:** \[ \frac{1}{\tan a}(\sin a \tan a + \cos a) \] 2. **Substitute \( \tan a = \frac{\sin a}{\cos a} \):** \[ = \frac{1}{\frac{\sin a}{\cos a}}(\sin a \cdot \frac{\sin a}{\cos a} + \cos a) \] \[ = \frac{\cos a}{\sin a} \left( \frac{\sin^2 a}{\cos a} + \cos a \right) \] 3. **Simplify the expression:** \[ = \frac{\cos a}{\sin a} \left( \frac{\sin^2 a + \cos^2 a \cdot \cos a}{\cos a} \right) \] \[ = \frac{\cos a}{\sin a} \left( \frac{1}{\cos a} \right) \quad \text{(since } \sin^2 a + \cos^2 a = 1\text{)} \] \[ = \frac{1}{\sin a} \] 4. **Thus, we have proven:** \[ \frac{1}{\tan a}(\sin a \tan a + \cos a) = \frac{1}{\sin a} \] ### Part 2: Solve \( (4 \theta - 8) \sin 30^{\circ} = \theta^{3} - 8 \) 1. **Substituting \( \sin 30^{\circ} = \frac{1}{2} \):** \[ (4 \theta - 8) \cdot \frac{1}{2} = \theta^{3} - 8 \] \[ 2 \theta - 4 = \theta^{3} - 8 \] 2. **Rearranging the equation:** \[ \theta^{3} - 2 \theta - 4 = 0 \] 3. **Using the Rational Root Theorem, we can test possible rational roots. Testing \( \theta = 2 \):** \[ 2^{3} - 2 \cdot 2 - 4 = 8 - 4 - 4 = 0 \] Thus, \( \theta = 2 \) is a root. 4. **Factoring out \( (\theta - 2) \):** \[ \theta^{3} - 2 \theta - 4 = (\theta - 2)(\theta^{2} + 2\theta + 2) \] 5. **Finding the roots of \( \theta^{2} + 2\theta + 2 = 0 \):** \[ \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i \] 6. **Thus, the solutions for \( \theta \) are:** \[ \theta = 2, \quad \theta = -1 + i, \quad \theta = -1 - i \] ### Part 3: Solve \( (Q^{2} + 2\theta + 4) = 2 \) 1. **Rearranging the equation:** \[ Q^{2} + 2\theta + 4 - 2 = 0 \] \[ Q^{2} + 2\theta + 2 = 0 \] 2. **This is a quadratic in \( Q \):** \[ Q^{2} = -2\theta - 2 \] ### Part 4: Determine \( \tan 240^{\circ} \) 1. **Using the reference angle:** \[ 240^{\circ} = 180^{\circ} + 60^{\circ} \] \[ \tan 240^{\circ} = \tan(180^{\circ} + 60^{\circ}) = \tan 60^{\circ} = \sqrt{3} \] Since \( 240^{\circ} \) is in the third quadrant, where tangent is positive: \[ \tan 240^{\circ} = \sqrt{3} \] ### Part 5: General Solutions a) **Solve \( 2 \cos 2\theta = -0.44 \):** \[ \cos 2\theta = -0.22 \] The general solution for \( \cos x = k \) is: \[ x = \pm \cos^{-1}(-0.22) + 2n\pi \] Thus, \[ 2\theta = \pm \cos^{-1}(-0.22) + 2n\pi \] \[ \theta = \frac{\pm \cos^{-1}(-0.22) + 2n\pi}{2} \] b) **Solve \( 2 \cos^{2} \theta - \cos Q = 0 \):** \[ 2 \cos^{2} \theta = \cos Q \] The general solution for \( \cos Q = k \) is: \[ Q = \pm \cos^{-1}(2 \cos^{2} \theta) + 2m\pi \] ### Summary of Solutions: - \( \theta = 2, -1 + i, -1 - i \) - \( \tan 240^{\circ} = \sqrt{3} \) - General solutions for \( 2 \cos 2\theta = -0.44 \) and \( 2 \cos^{2} \theta - \cos Q = 0 \) are provided above.