Originally Posted by
romsek you can note that
$\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$
and that
$\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$
so
$\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$
$\theta$ is obtuse so
$\sin(\theta) > 0$ and $\cos(\theta)<0$
so
$\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$
and thus
$\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$
for (9)
$\sec^2(\theta)=1+\tan^2(\theta)$
$\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$
so
$\tan(\theta)=a - \dfrac 1 {4a}$
$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$