Question

Given cos = -2/5 a in quadrant III and cos b = 1/4, b in quadrant I find

Answer 1

I guess you mean $\cos a=-\dfrac25$. Since $a$ is in quadrant III, we expect $\sin a$. Then

$\sin a=-\sqrt{1-\cos^2a}=-\dfrac{\sqrt{21}}5$

Since $b$ is in quadrant I, we expect $\sin b>0$, so that

$\sin b=\sqrt{1-\cos^2b}=\dfrac{\sqrt{15}}4$

Now,

$\sin(a+b)=\sin a\cos b+\cos a\sin b=-\dfrac{2\sqrt{15}+\sqrt{21}}{20}$

$\cos(a+b)=\cos a\cos b-\sin a\sin b=\dfrac{3\sqrt{35}-2}{20}$

and

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=-\dfrac{2\sqrt{15}+\sqrt{21}}{3\sqrt{35}-2}$