I'm trying to solve this problem but I can't figure how. Can you help me?

$$A=\frac{\sin \alpha+\cos(3\pi/2-\alpha)+\tan(5\pi+\alpha)}{\csc(2\pi-\alpha)+\sin(5\pi/2+\alpha)}$$

If $\tan \alpha=-2/3$ and $\alpha \in$ IV quadrant, calculate the value of the expression.


  • $\begingroup$ Pardon me, but what are "sen" and "tg"? Are these abbreviations for "sine" and "tangent"? if so, the only standard abbreviations in English are "sin" and "tan". $\endgroup$ – MJD Mar 26 '12 at 2:58
  • $\begingroup$ Fixed. Sorry for that, the english is not my native language. $\endgroup$ – ignaciotr Mar 26 '12 at 3:01
  • $\begingroup$ No problem; that's why I asked. $\endgroup$ – MJD Mar 26 '12 at 3:03
  • $\begingroup$ Can somebody help? $\endgroup$ – ignaciotr Mar 26 '12 at 23:33

You can find $\cos(\alpha)$ from the identity $\sec^2(\alpha) = \tan^2(\alpha) + 1$ and the fact that $\alpha$ is in QIV. Then find the sine. After that use angle sum/difference formulas to simplify terms that involve a trig function of a sum or difference. For example, $\cos(3\pi/2 - \alpha) = \cos(3\pi/2)\cos(\alpha) + \sin(3\pi/2)\sin(\alpha) = -\sin(\alpha)$. It's a bit messy but I don't see an easier solution to the problem.


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