$\begin{aligned} & \int_0^{\frac{\pi}{4}}\left(\tan ^{\mathrm{n}} x+\tan ^{\mathrm{n}-2} x\right) \mathrm{d} x=\frac{1}{\mathrm{n}-1} \\ & \therefore \quad \mathrm{I}_{12}+\mathrm{I}_{10}=\int_0^{\frac{\pi}{4}}\left(\tan ^{12} \theta+\tan ^{10} \theta\right) \mathrm{d} \theta \\ &=\frac{1}{12-1} \\ &=\frac{1}{11}\end{aligned}$