in terms of sin(a) or cos() (or both) only. In some cases, it might be helpful to make a little sketch of the situation,either on paper or in your mind, to help you solving the exercises. To illustrate what I expect you to do, the first two parts are already solved, so you need to start the exercise by solving (c). \begin{aligned} &\text { (a) } \sec \left(\frac{\pi}{2}-\alpha\right) \text { . Solution. We know that } \sec \left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\cos \left(\frac{\pi}{2}-\alpha\right)} . \text { Since sine and cosine are cofunc- }\\ &\text { tions, we also know that } \cos \left(\frac{\pi}{2}-\alpha\right)=\sin (\alpha) . \text { Therefore, } \sec \left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\sin (\alpha)} \text { . } \end{aligned} \text { (c) } \cot \left(\frac{\pi}{2}-\alpha\right) \text { . } \text { (d) } \tan (\pi+\alpha) \text { . } \text { (b) } \cos (\pi-\alpha) \text { . Solution. The angle } \beta=\pi-\alpha \text { is on the second quadrant, and its reference angle is } \alpha \text { . } \text { Thus, } \cos (\pi-\alpha)=-\cos (\alpha) \text { (the negative sign is because of } \pi-\alpha \text { being on the second quadrant). } \text { (e) } \tan (\pi-\alpha) \text { . } \text { (g) } \cos (2 \pi+\alpha) \text { (f) } \sin (2 \pi-\alpha) \text { . } \text { (h) } \cos (\alpha-2 \pi) \text { . }

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