Assertion (A) : The angle between the curves y=sinx, y=cosx is ${\tan }^{-1}\sqrt{2}$
Reason (R) : The slopes of their tangents at $\left(\frac{\mathrm{\pi }}{4},\frac{1}{\sqrt{2}}\right)$ are $\frac{1}{\sqrt{2}}$ and $\frac{-1}{\sqrt{2}}$  respectively
 

$\mathrm{y}=\mathrm{sinx}-\left(1\right) \mathrm{y}=\mathrm{cosx}-\left(2\right)$
Solving $\mathrm{sinx}=\mathrm{cotx}\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{4}$
$\Rightarrow \mathrm{y}=\frac{1}{\sqrt{2}}$
$\therefore \mathrm{P}.\mathrm{I}=\left(\frac{\mathrm{\pi }}{4},\frac{1}{\sqrt{2}}\right)$
Diff (1) w.r.t. ‘x’
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x}\\ {\mathrm{m}}_1={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=\frac{\mathrm{\pi }}{4}}=\frac{1}{\sqrt{2}}\end{array}$
Diff (2) w.r.t. ‘x’
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x}\\ {\mathrm{m}}_2={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=\mathrm{\pi }/4}=-\frac{1}{\sqrt{2}}\\ {\mathrm{m}}_1=\frac{1}{\sqrt{2}},{\mathrm{m}}_2=-\frac{1}{\sqrt{2}}\end{array}$
$\begin{array}{l}\mathrm{Tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ =\left|\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1-\frac{1}{2}}\right|\\ =\left|\frac{\sqrt{2}}{1/2}\right|\\ =2\sqrt{2}\\ \mathrm{\theta }={\mathrm{Tan}}^{-1}\left(2\sqrt{2}\right)\end{array}$
$\therefore$A is false, R is true