Trigonometric Functions, Identities and Equations

Multiply by $\dfrac{\pi}{180}$: $135 \times \dfrac{\pi}{180} = \dfrac{135\pi}{180}$, and $\tfrac{135}{180}$ cancels to $\tfrac34$.

The sine curve crosses the axis at the start, the halfway point and the end of a full turn. All three lie in the interval, so all three count.

$\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$, so $\cos\theta = \tfrac45$ — positive, because $\theta$ is acute. Then $\tan\theta = \sin\theta \div \cos\theta = \tfrac35 \div \tfrac45 = \tfrac34$.

$\cos\theta = -\tfrac12$. The related acute angle is $60°$, and cosine is negative in the second and third quadrants, giving $180° - 60° = 120°$ and $180° + 60° = 240°$.

Replace $\sin^2\theta$ with $1 - \cos^2\theta$: $2 - 2\cos^2\theta + 3\cos\theta = 3$, so $2\cos^2\theta - 3\cos\theta + 1 = 0$, that is $(2\cos\theta - 1)(\cos\theta - 1) = 0$. Then $\cos\theta = \tfrac12$ gives $60°$ and $300°$, and $\cos\theta = 1$ gives $0°$ and $360°$.

Over the common denominator $\sin\theta(1+\cos\theta)$ the numerator is $\sin^2\theta + (1+\cos\theta)^2 = \sin^2\theta + 1 + 2\cos\theta + \cos^2\theta$. Since $\sin^2\theta + \cos^2\theta = 1$, that is $2 + 2\cos\theta = 2(1+\cos\theta)$. Cancelling the $(1+\cos\theta)$ leaves $\dfrac{2}{\sin\theta}$.