(1)

$ $$\begin{eqnarray*} \sin 900^\circ & = & \sin \left( 180^\circ + 360^\circ \cdot 2 \right)\\[0.5em] & = & \sin 180^\circ = 0\end{eqnarray*}$$ $

(2)

$ $$\begin{eqnarray*} \sin \left( - 600^\circ\right) & = & \sin \left( 120^\circ - 360^\circ \cdot 2 \right)\\[0.5em] & = & \sin 120^\circ = \dfrac{\sqrt{3}}{2}\end{eqnarray*}$$ $

(3)

$ $$\begin{eqnarray*} \cos 585^\circ & = & \cos \left( 225^\circ + 360^\circ \right)\\[0.5em] & = & \cos 225^\circ = -\dfrac{\sqrt{2}}{2}\end{eqnarray*}$$ $

(4)

$ $$\begin{eqnarray*} \cos \left( -1050^\circ \right) & = & \cos \left( 30^\circ - 360^\circ \cdot 3 \right)\\[0.5em] & = & \cos 30^\circ = \dfrac{\sqrt{3}}{2}\end{eqnarray*}$$ $

(4)

$ $$\begin{eqnarray*} \tan 945^\circ & = & \tan \left( 225^\circ + 360^\circ \cdot 2 \right)\\[0.5em] & = & \tan 225^\circ = 1\end{eqnarray*}$$ $

(6)

$ $$\begin{eqnarray*} \tan \left( -510^\circ \right) & = & \tan \left( 210^\circ - 360^\circ \cdot 2 \right)\\[0.5em] & = & \tan 210^\circ = \dfrac{\sqrt{3}}{3}\end{eqnarray*}$$ $

(1)

$\dfrac{\sin \theta}{1 + \cos \theta} = \dfrac{\sin \theta(1 - \cos \theta)}{1-\cos^2 \theta} = \dfrac{\sin \theta(1-\cos \theta)}{\sin^2 \theta} = \dfrac{1-\cos \theta}{\sin \theta}$

より

$\dfrac{\sin \theta}{1+\cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = \dfrac{1- \cos \theta}{\sin \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = \dfrac{2}{\sin \theta}$

(2)

$ $$\begin{eqnarray*} \dfrac{1}{\tan \theta} + \dfrac{1}{\sin \theta} & = & \dfrac{\cos \theta}{\sin \theta} + \dfrac{1}{\sin \theta}\\[1em] & = & \dfrac{1 + \cos \theta}{\sin \theta}\\[1em] & = & \dfrac{1- \cos^2 \theta}{\sin \theta(1- \cos \theta)}\\[1em] & = & \dfrac{\sin^2 \theta}{\sin \theta(1-\cos \theta)} = \dfrac{\sin \theta}{1- \cos \theta}\end{eqnarray*}$$ $

(3)

$ $$\begin{eqnarray*}\sin^2 \theta + \cos^4 \theta & = & \sin^2 \theta + (1-\sin^2 \theta)^2\\[0.5em] & = & \sin^2 \theta + \left( 1-2 \sin^2 \theta + \sin^4 \theta \right)\\[0.5em] & = & (1- \sin^2 \theta) + \sin^4 \theta = \cos^2 \theta + \sin^4 \theta\end{eqnarray*}$$ $

(4)

$\dfrac{\tan \theta}{\sin \theta} = \dfrac{\sin \theta}{\cos \theta}\cdot \dfrac{1}{\sin \theta} = \dfrac{1}{\cos \theta}$

より

$ $$\begin{eqnarray*}\dfrac{\tan \theta}{\sin \theta} - \dfrac{\sin \theta}{\tan \theta} & = & \dfrac{1}{\cos \theta} - \cos \theta\\[0.5em] & = & \dfrac{1 - \cos^2 \theta}{\cos \theta}\\[0.5em] & = & \dfrac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \dfrac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta\end{eqnarray*}$$ $