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Find the value of the trigonometric function $\,\sec(v-u)\,$ given that

$$\sin u=−\frac{20}{29}\;\;,\;\; \cos v =−\frac{3}{5}$$

(Both u and v are in Quadrant III.)


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NO! First wait until you understand how to do the other problem you have posted, which is very, very much like this one. Then, when you have digested that other problem, try this one again --- I bet you'll be able to do it! And, if you can't, that's the time to come back and ask for clarification! – Gerry Myerson Oct 29 '12 at 2:32

$$\sec(v-u)=\frac{1}{\cos(v-u)}=\frac{1}{\cos v\cos u+\sin v\sin u}=$$

$$\frac{1}{-\frac{3}{5}\cos u-\frac{20}{29}\sin v}$$

But since $\,u,v\,$ in the third quadrant then

$$\cos u=-\sqrt{1-\sin^2u}=-\sqrt{1-\frac{20^2}{29^2}}=-\frac{21}{29}$$

$$\sin v=-\sqrt{1-\cos^2v} \,...\,etc.$$

End the exercise now.

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So we want $\dfrac{1}{\cos(v-u)}$. We will be nearly finished once we know $\cos(v-u)$. By a standard formula, $$\cos(v-u)=\cos v\cos u+\sin v\sin u.$$

To find $\cos u$, note that we are in the third quadrant so $\cos u$ is negative. We have $\cos^2 u+\sin^2 u=1$. So $\cos^2u=1-\frac{400}{841}=\frac{441}{841}$ and therefore $\cos u=-\frac{21}{29}$. Finding $\cos v$ is similar, but the arithmetic is simpler.

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Use the ID:

$$\cos(v-u)=\cos(v)\cos(v)+\sin(u)\sin(u)$$ $$\frac{1}{\cos(v-u)}=\frac{1}{\cos(v)\cos(v)+\sin(u)\sin(u)}$$ $$\sec(v-u)=\frac{1}{\cos(v)\cos(v)+\sin(u)\sin(u)}$$

Now, use the Pythagorean thm and quadrant number to find $\cos(u)$ and $\sin(v)$:

$$\sin(u) = \frac y r$$ $$x = -\sqrt{r^2-y^2}$$ $$\cos(u) = \frac{-\sqrt{r^2-y^2}}{r}$$ $$\cos(u) = \frac{-\sqrt{29^2-20^2}}{29}$$

$\sin(v)$ is similar, and so omitted.

Now plug in to the secant formula.

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