Here are the solutions to the questions: i) Express $-150^\circ$ into radians. Step 1: Use the conversion factor $\frac{\pi \text{ radians}}{180^\circ}$ to convert degrees to radians. $$ -150^\circ \times \frac{\pi}{180^\circ} $$ Step 2: Simplify the fraction. $$ -\frac{150\pi}{180} = -\frac{15\pi}{18} = -\frac{5\pi}{6} $$ The angle in radians is $\boxed{-\frac{5\pi}{6} \text{ radians}}$. ii) Find $\theta$, when $l=2\text{cm}$ and $r=3.5\text{cm}$. Step 1: Use the formula for arc length, $l = r\theta$, where $\theta$ is in radians. Rearrange the formula to solve for $\theta$: $$ \theta = \frac{l}{r} $$ Step 2: Substitute the given values for $l$ and $r$. $$ \theta = \frac{2\text{ cm}}{3.5\text{ cm}} $$ Step 3: Calculate the value of $\theta$. $$ \theta = \frac{2}{3.5} = \frac{2}{\frac{7}{2}} = \frac{4}{7} $$ The angle $\theta$ is $\boxed{\frac{4}{7} \text{ radians}}$. iii) Verify that: $\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta$. Step 1: Start with the left-hand side (LHS) of the equation. $$ \text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta $$ Step 2: Find a common denominator for the terms, which is $\cos \theta$. $$ \text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos \theta \cdot \cos \theta}{\cos \theta} $$ $$ \text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta} $$ Step 3: Combine the terms over the common denominator. $$ \text{LHS} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta} $$ Step 4: Apply the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$. $$ \text{LHS} = \frac{1}{\cos \theta} $$ Step 5: Recognize that $\frac{1}{\cos \theta}$ is equal to $\sec \theta$. $$ \text{LHS} = \sec \theta $$ This matches the right-hand side (RHS) of the equation. Therefore, the identity is verified. $\boxed{\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta}$ That's 2 down. 3 left today — send the next one.