How do you evaluate the expression tan(u-v) given sinu=3/5 with pi/2<u<p and cosv=-5/6 with pi<v<(3pi)/2?

How do you evaluate the expression $\tan \left(u - v\right)$ given $\sin u = \frac{3}{5}$ with $\frac{\pi}{2} < u < \pi$ and $\cos v = - \frac{5}{6}$ with $\pi < v < \frac{3 \pi}{2}$?

Aug 5, 2018

$\tan \left(u - v\right) = - \frac{432 + 125 \sqrt{11}}{301}$
$\tan \left(u - v\right) \approx - 2.8126$

Explanation:

Here ,

$\sin u = \frac{3}{5} > 0 , w i t h , \frac{\pi}{2} < u < \pi \to {2}^{n d} Q u a \mathrm{dr} a n t$

$\therefore \cos u = - \sqrt{1 - {\sin}^{2} u} = - \sqrt{1 - \frac{9}{25}} = - \sqrt{\frac{16}{25}}$

$\therefore \cos u = - \frac{4}{5}$

$\therefore \tan u = \sin \frac{u}{\cos} u = \left(\frac{\frac{3}{5}}{- \frac{4}{5}}\right)$

:.color(blue)(tanu=-3/4

Again ,

$\cos v = - \frac{5}{6} < 0 , w i t h , \pi < v < \frac{3 \pi}{2} \to {3}^{r d} Q u a \mathrm{dr} a n t$

$\therefore \sin v = - \sqrt{1 - {\cos}^{2} v} = - \sqrt{1 - \frac{25}{36}} = - \frac{\sqrt{11}}{6}$

$\therefore \tan v = \sin \frac{v}{\cos} v = \left(\frac{- \frac{\sqrt{11}}{6}}{- \frac{5}{6}}\right)$

:.color(blue)(tanv=sqrt11/5

So ,

$\tan \left(u - v\right) = \frac{\tan u - \tan v}{1 + \tan u \tan v}$

color(white)(tan(u-v))=(-3/4-sqrt11/5)/(1+(-3/4)(sqrt11/5)

color(white)(tan(u-v))=(-15-4sqrt11)/(20-3sqrt11

$\textcolor{w h i t e}{\tan \left(u - v\right)} = \frac{- 15 - 4 \sqrt{11}}{20 - 3 \sqrt{11}} \times \frac{20 + 3 \sqrt{11}}{20 + 3 \sqrt{11}}$

$\textcolor{w h i t e}{\tan \left(u - v\right)} = \frac{- 300 - 45 \sqrt{11} - 80 \sqrt{11} - 132}{{\left(20\right)}^{2} - {\left(3 \sqrt{11}\right)}^{2}}$

$\textcolor{w h i t e}{\tan \left(u - v\right)} = \frac{- 432 - 125 \sqrt{11}}{400 - 99}$

$\tan \left(u - v\right) = - \frac{432 + 125 \sqrt{11}}{301}$

$\tan \left(u - v\right) \approx - 2.8126$