1. We are given that $\tan x = \frac{1}{\sqrt{3}}$ with $0^\circ \leq x \leq 90^\circ$. We want to find $\cos x - \sin x$. 2. Recall that $\tan x = \frac{\sin x}{\cos x}$. So, $\frac{\sin x}{\cos x} = \frac{1}{\sqrt{3}}$. 3. From this, $\sin x = \frac{\cos x}{\sqrt{3}}$. 4. Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, substitute $\sin x$: $$\left(\frac{\cos x}{\sqrt{3}}\right)^2 + \cos^2 x = 1$$ $$\frac{\cos^2 x}{3} + \cos^2 x = 1$$ $$\frac{\cos^2 x}{3} + \frac{3\cos^2 x}{3} = 1$$ $$\frac{4\cos^2 x}{3} = 1$$ 5. Multiply both sides by 3: $$4\cos^2 x = 3$$ 6. Divide both sides by 4: $$\cos^2 x = \frac{3}{4}$$ 7. Taking the positive root (since $0^\circ \leq x \leq 90^\circ$, cosine is positive): $$\cos x = \frac{\sqrt{3}}{2}$$ 8. Recall from step 3: $$\sin x = \frac{\cos x}{\sqrt{3}} = \frac{\frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{1}{2}$$ 9. Calculate $\cos x - \sin x$: $$\frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$$ 10. Approximate using $\sqrt{3} = 1.732$: $$\frac{1.732 - 1}{2} = \frac{0.732}{2} = 0.366$$ **Final answer:** $$\cos x - \sin x = \frac{\sqrt{3} - 1}{2} \approx 0.366$$ Next, evaluate the expressions involving surds given $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$: **a) Evaluate $\frac{1}{\sqrt{3}}$** 1. Substitute $\sqrt{3} = 1.732$: $$\frac{1}{1.732} \approx 0.577$$ **b) Evaluate $\frac{2}{\sqrt{2}}$** 1. Substitute $\sqrt{2} = 1.414$: $$\frac{2}{1.414} \approx 1.414$$ **c) Evaluate $\frac{2}{\sqrt{3} + \sqrt{2}}$** 1. Substitute values: $$\frac{2}{1.732 + 1.414} = \frac{2}{3.146} \approx 0.636$$ **d) Evaluate $\frac{1}{\sqrt{3} - \sqrt{2}}$** 1. Substitute values: $$\frac{1}{1.732 - 1.414} = \frac{1}{0.318} \approx 3.145$$ These values are rounded to 3 significant figures as requested.