1. The problem asks to find other trigonometric function values given one function value and the interval for $x$. 2. We will solve part (a) only, as per instructions to solve the first question only. 3. Given: $\frac{\pi}{2} < x < 0$ and $\sin x = -0.6$. 4. Note that the interval $\frac{\pi}{2} < x < 0$ is unusual since $\frac{\pi}{2} \approx 1.57$ and $0$ is less than $\frac{\pi}{2}$. Assuming the interval means $0 < x < \frac{\pi}{2}$ or $-\frac{\pi}{2} < x < 0$ (likely a typo), we proceed with $-\frac{\pi}{2} < x < 0$ because $\sin x$ is negative there. 5. Use the Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$ 6. Substitute $\sin x = -0.6$: $$(-0.6)^2 + \cos^2 x = 1$$ $$0.36 + \cos^2 x = 1$$ 7. Solve for $\cos^2 x$: $$\cos^2 x = 1 - 0.36 = 0.64$$ 8. Take the square root: $$\cos x = \pm \sqrt{0.64} = \pm 0.8$$ 9. Since $x$ is in the interval $-\frac{\pi}{2} < x < 0$, cosine is positive (cosine is positive in the fourth quadrant), so: $$\cos x = 0.8$$ 10. Calculate $\tan x$ using: $$\tan x = \frac{\sin x}{\cos x} = \frac{-0.6}{0.8} = -0.75$$ 11. Calculate $\cot x$ as reciprocal of $\tan x$: $$\cot x = \frac{1}{\tan x} = \frac{1}{-0.75} = -\frac{4}{3}$$ 12. Calculate $\sec x$ as reciprocal of $\cos x$: $$\sec x = \frac{1}{0.8} = 1.25$$ 13. Calculate $\csc x$ as reciprocal of $\sin x$: $$\csc x = \frac{1}{-0.6} = -\frac{5}{3}$$ Final answers: $$\cos x = 0.8$$ $$\tan x = -0.75$$ $$\cot x = -\frac{4}{3}$$ $$\sec x = 1.25$$ $$\csc x = -\frac{5}{3}$$