Equivalent to $$\sum_{1\le i<j\le n}(\tan\theta_{i}\cot\theta_{j}+\tan\theta_{j}\cot\theta_{i})\ge2 \sum_{1\le i<j\le n}(\sin\theta_{i}\sin\theta_{j}+\cos\theta_{i}\cos\theta_{j})$$We'll prove that for all integer $1\le i<j\le n$ holds $$\tan\theta_{i}\cot\theta_{j}+\tan\theta_{j}\cot\theta_{i}\ge 2(\sin\theta_{i}\sin\theta_{j}+\cos\theta_{i}\cos\theta_{j})$$Let's perform a substitution to make it clear $$a=\tan\theta_{i},\ b=\tan\theta_{j}, (a,b)\in R_+^2$$$$\frac{a}{b}+\frac{b}{a}\ge 2\left(\frac{a}{\sqrt{a^2+1}}\cdot\frac{b}{\sqrt{b^2+1}}+\frac{1}{\sqrt{a^2+1}}\cdot\frac{1}{\sqrt{b^2+1}}\right)$$This is clear, because by AM-GM and C-S we have $$\frac{a}{b}+\frac{b}{a}\ge 2$$and $$\sqrt{(a^2+1)(b^2+1)}\ge ab+1$$This proves our inequality