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) \( 2 \cos ^{2} \theta-\cos Q=0 \)

To prove that \( \frac{1}{\tan a} (\sin a \tan a + \cos a) = \frac{1}{\sin a} \), we can start by rewriting \(\tan a\) as \(\frac{\sin a}{\cos a}\). Thus, our left side becomes: \[ \frac{1}{\frac{\sin a}{\cos a}} \left( \sin a \cdot \frac{\sin a}{\cos a} + \cos a \right) = \frac{\cos a}{\sin a} \left( \frac{\sin^2 a}{\cos a} + \cos a \right). \] This simplifies to: \[ \frac{\cos a}{\sin a} \cdot \left( \frac{\sin^2 a + \cos^2 a \cdot \cos a}{\cos a} \right) = \frac{\cos a}{\sin a} \cdot \left( \frac{1}{\cos a} \right) = \frac{1}{\sin a}. \] Now it's proven! Next, regarding the equations given, first, let's solve \((4\theta - 8) \sin 30^{\circ} = \theta^3 - 8\). Since \(\sin 30^{\circ} = \frac{1}{2}\): \[ (4\theta - 8) \cdot \frac{1}{2} = \theta^3 - 8 \implies 2\theta - 4 = \theta^3 - 8. \] Rearranging gives: \[ \theta^3 - 2\theta - 4 = 0. \] Using synthetic division or testing roots can help solve this cubic equation, but let's save that for later exploration. For the quadratic equation \((Q^2 + 2\theta + 4 = 2)\): \[ Q^2 + 2\theta + 2 = 0 \implies Q^2 = -2\theta - 2. \] Notably, \(\tan(240^\circ) = \tan(180^\circ + 60^\circ) = \tan 60^\circ = \sqrt{3}\), but negative since \(240^\circ\) is in the third quadrant, giving \(\tan(240^\circ) = \sqrt{3}\). Now, regarding the general solutions to the equations: 1. For \(2\cos 2\theta = -0.44\), we have \(\cos 2\theta = -0.22\). The angles that solve this can be found using \(\cos 2\theta = x\) where \(x = -0.22\); the general solution will be: \[ 2\theta = \pm \arccos(-0.22) + 2k\pi. \] Thus, \(\theta = \frac{\pm \arccos(-0.22)}{2} + k\pi\). 2. For \(2\cos^2 \theta - \cos Q = 0\): This gives \(\cos^2 \theta = \frac{1}{2}\) which leads to \(\cos \theta = \pm \frac{1}{\sqrt{2}}\). The solutions are: \[ \theta = \frac{\pi}{4} + k\pi \quad \text{and} \quad \theta = \frac{3\pi}{4} + k\pi. \] Put those values in your math toolbox and you'll be navigating trigonometric equations like a pro!