Can someone help me solve this:
tan(a-b)
cosa = -4/5 Lies in Quadrant 3
sinb = 1/3 Lies in Quadrant 1
for angle $\displaystyle a$ , sketch a reference triangle in quad III
adjacent side = -4 , hypotenuse = 5 , opposite side = ?
$\displaystyle \tan{a} = \frac{opp}{adj}$
for angle b , sketch a reference triangle in quad I
opposite side = 1 , hypotenuse = 3 , adjacent side = ?
$\displaystyle \tan{b} = \frac{opp}{adj}$
lastly, calculate $\displaystyle \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$
note that $\displaystyle \frac{\sqrt{8}}{8} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}$
$\displaystyle \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$
$\displaystyle \tan(a-b)=\frac{\frac{3}{4}-\frac{\sqrt{2}}{4}}{1+\frac{3\sqrt{2}}{16}}$
multiply by $\displaystyle \frac{16}{16}$ to clear the fractions ...
$\displaystyle \tan(a-b) = \frac{4(3-\sqrt{2})}{16+3\sqrt{2}}$