The following is an attempted proof of the conjecture:

if $\tan \theta >0$, then $\sin \theta +\cos \theta >1$.

Suppose $\tan \theta >0$, so in particular $\cos \theta \ne 0$.

Since $\tan \theta =\frac{\sin \theta }{\cos \theta }$, then $\sin \theta \cos \theta =\tan \theta {\cos }^2\theta >0$. (I)

It follows that $1+2\sin \theta \cos \theta >1$. (II)

Therefore ${\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta >1$, (III)

which factorises to give $(\sin \theta +\cos \theta {)}^2>1$. (IV)

Therefore $\sin \theta +\cos \theta >1$. (V)

Which one of the following is the case?