The following is an attempted proof of the conjecture:
$$\text { if } \tan \theta > 0 , \text { then } \sin \theta + \cos \theta > 1$$
Suppose $\tan \theta > 0$, so in particular $\cos \theta \neq 0$.
$$\text { Since } \tan \theta = \frac { \sin \theta } { \cos \theta } \text {, then } \sin \theta \cos \theta = \tan \theta \cos ^ { 2 } \theta > 0 . $$
It follows that $1 + 2 \sin \theta \cos \theta > 1$.
Therefore $\sin ^ { 2 } \theta + 2 \sin \theta \cos \theta + \cos ^ { 2 } \theta > 1$,
which factorises to give $( \sin \theta + \cos \theta ) ^ { 2 } > 1$.
Therefore $\sin \theta + \cos \theta > 1$.
Which one of the following is the case?