Hello there. To solve this question, we'll have to remember some properties about trigonometric functions and sum of angles.
Given the following functions, we want to express them as a function of an acute angle:
[tex]\begin{gathered} \tan(110^{\circ}) \\ \\ \cos(200^{\circ}) \\ \end{gathered}[/tex]First, remember the following properties:
[tex]\begin{gathered} \tan(\alpha+90^{\circ})=-\cot(\alpha) \\ \\ \cos(\alpha+\pi)=-\cos(\alpha) \\ \end{gathered}[/tex]That we can prove easily knowing some properties of sum of angles:
[tex]\begin{gathered} \tan(\alpha+90^{\circ})=\dfrac{\sin(\alpha+90^{\circ})}{\cos(\alpha+90^{\circ})} \\ \\ \text{ Whereas }\sin(\alpha+90^{\circ})=\cos(\alpha)\text{ and }\cos(\alpha+90^{\circ})=-\sin(\alpha) \\ \\ \text{ Therefore} \\ \\ \tan(\alpha+90^{\circ})=\dfrac{\cos(\alpha)}{-\sin(\alpha)}=-\cot(\alpha)_{\text{ }\square} \\ \\ \\ \end{gathered}[/tex]For the second, we use the same formula for sum of angles for cosines
[tex]\begin{gathered} \cos(\alpha+180^{\circ})=\cos(\alpha)\cos(180^{\circ})-\sin(\alpha)\sin(180^{\circ}) \\ \\ \text{ But }\cos(180^{\circ})=-1\text{ and }\sin(180^{\circ})=0 \\ \\ \text{ Hence} \\ \\ \cos(\alpha+180^{\circ})=-\cos(\alpha)\text{ }_{\square} \\ \\ \\ \\ \\ \end{gathered}[/tex]In this case, notice that
[tex]\tan(110^{\circ})=\tan(\alpha+90^{\circ})[/tex]It is easy to see that
[tex]\alpha=20^{\circ}[/tex]Such that the function expressing this as a function of an acute angle is
[tex]-\cot(20^{\circ})[/tex]And for the cosine, we'll get
[tex]\cos(200^{\circ})=\cos(\beta+180^{\circ})[/tex]Again we see that
[tex]\beta=20^{\circ}[/tex]Such that we get the function
[tex]-\cos(20^{\circ})[/tex]There are the answer to this question.
