Statement (I):Minimum value of  $(3\sin \theta -4\cos \theta )(3\cos \theta +4\sin \theta )\text{ is }-\frac{25}{2}$

Statement (II): Minimum value of $9{\tan }^{2}x+16{\cot }^{2}x\text{ is }24$

Which of the above statements is correct ?

$$\begin{gathered} \mathrm{Statement} \left(\mathrm{I}\right):(3\sin \mathrm{\theta }-4\cos \mathrm{\theta })(3\cos \mathrm{\theta }+4\sin \mathrm{\theta }) \\ =25(\frac{3}{5}\sin \mathrm{\theta }-\frac{4}{5}\cos \mathrm{\theta })(\frac{3}{5}\cos \mathrm{\theta }+\frac{4}{5}\sin \mathrm{\theta }) \\ \mathrm{Let} \mathrm{cos\alpha }=\frac{3}{5},\mathrm{sin\alpha }=\frac{4}{5} \\ (3\sin \mathrm{\theta }-4\cos \mathrm{\theta })(3\cos \mathrm{\theta }+4\sin \mathrm{\theta })=25\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)\cos \left(\mathrm{\theta }-\mathrm{\alpha }\right) \\ =\frac{25}{2}\sin 2\left(\mathrm{\theta }-\mathrm{\alpha }\right) \\ \therefore \min \mathrm{value}=\frac{-25}{2} \end{gathered}$$

Statement (II)$$\begin{gathered} \frac{9{\tan }^{2}x+16co{t}^{2}x}{2}\ge \sqrt{\left(9{\tan }^{2}x\right)\left(16co{t}^{2}x\right)} \\ \Rightarrow 9{\tan }^{2}x+16co{t}^{2}x\ge 2\left(3\right)\left(4\right) \\ \therefore 9{\tan }^{2}x+16co{t}^{2}x\ge 24 \end{gathered}$$