The angle of intersection of the curves

Given that the curve $y=2 \sin ^{2} x$ and $y=\cos ^{2} x$

Differentiating both w.r.t. $x$,

$\frac{d y}{d x}=4 \sin x \cos x$ and $\frac{d y}{d x}=-2 \cos x \sin x$

$\mathrm{m}_{1}=4 \sin \mathrm{x} \cos \mathrm{x}$ and $\mathrm{m}_{2}=-2 \cos \mathrm{x} \sin \mathrm{x}$

$\operatorname{At} \mathrm{X}=\frac{\pi}{6}$

$\mathrm{m}_{1}=\sqrt{3}$ and $\mathrm{m}_{2}=-\frac{\sqrt{3}}{2}$

$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\sqrt{3}+\frac{\sqrt{3}}{2}}{1-\sqrt{3} \times \frac{\sqrt{3}}{2}}\right|=\frac{\frac{3 \sqrt{3}}{2}}{\frac{1}{2}}=3 \sqrt{3}$

$\Rightarrow \theta=\tan ^{-1} 3 \sqrt{3}$