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

$\tan(\theta)=a - \dfrac 1 {4a}$

$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$