Hint: Let $ x = \cos \theta$, $y = \cos \phi$.
Show by expansion (and clearing denominators) that both equations are equivalent to $x^4 - 2x^2 y^2 + y^4 =0$, hence these statements are equivalent to each other.
Note: This shows that the condition is satisfied iff $x = \pm y$. This is not required, but very strongly hinted at in the question.
You are on the right track.
writing $\tan\theta$ as$ \dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $ \dfrac {\cos\theta}{\sin\theta} $, we get
$ \dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$
$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)
$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$
$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)
$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)
which is
$1+\sec\theta\cdot\csc\theta$
QED.
Best Answer
$$\frac{\sin(x+\theta)}{\sin(x+\phi)}=\frac{\sin[(x+\phi)+(\theta-\phi)]}{\sin(x+\phi)}=\frac{\sin(x+\phi)\cos(\theta-\phi)+\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\\ =\frac{\sin(x+\phi)\cos(\theta-\phi)}{\sin(x+\phi)}+\frac{\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\cos(\theta-\phi)+\cot(x+\phi)\sin(\theta-\phi)$$
for the last equality I used $\cot(x+\phi)=\frac{\cos(x+\phi)}{\sin (x+\phi)}$