<s>[INST] Find the exact value of $\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha)$ Given that $\tan(\alpha) = -3$ find the exact value.
I tried:
$$\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha) = \\
(\sin(\pi)\cos(\alpha)+\cos(\pi)\sin(\alpha))+(\cos(\frac{3\pi}{2})\cos(\alpha)-\sin(\frac{3\pi}{2})\sin(\alpha))+\tan(-\frac{\pi}{2}+\alpha) = \\
-\sin(\alpha)+\sin(\alpha)+\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-(-\tan(\frac{\pi}{2})(-3)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-3\tan(\frac{\pi}{2})} = \\
???$$
What do I do next? 
 [/INST] You are right, the first two terms cancel out. However, the third term cannot be evaluated using the known formula, since both numerator and denominator are infinite.
Instead, consider the following chain of statements:
$$
\tan \left(-\frac \pi 2 + \alpha \right) = - \tan\left(\frac \pi 2 - \alpha\right) = -\cot \alpha = \frac{-1}{\tan \alpha} = \frac{-1}{3}
$$
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