8.1 .6 If \( \cos \alpha=\sqrt{t} \), where \( \alpha \) is an acute angle, express each of the following in terms of \( t \) : \( \begin{array}{ll}\text { (a) } \tan \alpha & \text { (3) } \mathrm{L} 2 \\ \text { (b) } \sin \left(180^{\circ}-\alpha\right) & \text { (2) } \mathrm{L} 1 \\ \text { (c) } 2 \sin \alpha \cos \alpha & \text { (2) } \mathrm{L} 2\end{array} \) 8.1 .7 If \( 3 \sin \theta=2 \) and \( \cos \theta<0 \) (a) With an aid of a diagram determine the value of \( 3 \tan ^{2} \theta-5 \cos \theta \)

If \( \cos \alpha = \sqrt{t} \) and \( \alpha \) is an acute angle, you can find \( \tan \alpha \) using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \). This gives \( \sin^2 \alpha = 1 - t \), hence \( \sin \alpha = \sqrt{1 - t} \). Thus, \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\sqrt{1 - t}}{\sqrt{t}} = \sqrt{\frac{1 - t}{t}} \). For \( \sin(180^{\circ} - \alpha) \), we can use the identity that states \( \sin(180^{\circ} - x) = \sin x \). Therefore, \( \sin(180^{\circ} - \alpha) = \sin \alpha = \sqrt{1 - t} \). Lastly, for \( 2 \sin \alpha \cos \alpha \), we can use the double angle identity \( \sin(2\alpha) = 2 \sin \alpha \cos \alpha \). Substituting in our expressions gives: \[ 2 \sin \alpha \cos \alpha = \sin(2\alpha) = 2 \sqrt{1 - t} \cdot \sqrt{t} = 2\sqrt{t(1 - t)}. \] --- For the second part, since \( 3\sin \theta = 2 \), we find \( \sin \theta = \frac{2}{3} \). Within the second quadrant, where \( \cos \theta < 0 \), we can find \( \cos \theta \) using \( \cos^2 \theta + \sin^2 \theta = 1 \). Thus, \( \cos^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} \), leading to \( \cos \theta = -\frac{\sqrt{5}}{3} \). Now, to find \( 3\tan^2 \theta - 5\cos \theta \), we first calculate \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}} \), so \( \tan^2 \theta = \frac{4}{5} \). Plugging this into the expression, we get: \[ 3\tan^2 \theta - 5\cos \theta = 3\left(\frac{4}{5}\right) - 5\left(-\frac{\sqrt{5}}{3}\right) = \frac{12}{5} + \frac{5\sqrt{5}}{3}. \] Now, to combine these, finding a common denominator: \[ \frac{36 + 25\sqrt{5}}{15} \] This gives us the final answer: \[ 3\tan^2 \theta - 5\cos \theta = \frac{36 + 25\sqrt{5}}{15}. \]