Explanation

(a)
Equation of given curves are
$y = \sin x$ …(i)
and $y = \cos x$ …(ii)
On solving Eqs. (i) and (ii), we get
$x = \frac{\pi}{4}$
$\therefore$ Point of intersection of curves is $\left( \frac{\pi}{4},\frac{1}{\sqrt{2}} \right)$
For $y = \sin x,\frac{dy}{dx} = \cos x$
$\Rightarrow \ \ \ \left( \frac{dy}{dx} \right)_{x = \pi/4} = \frac{1}{\sqrt{2}} = m_{1}$ (say)
For $y = \cos x\ \ \ \Rightarrow \ \frac{dy}{dx} = - \sin x$
$\Rightarrow \ \ \ \left( \frac{dy}{dx} \right)_{x = \pi/4} = - \frac{1}{\sqrt{2}} = m_{2}$ (say)
$\therefore\ \tan\theta = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}} = \frac{\frac{2}{\sqrt{2}}}{\frac{1}{2}}$
$\Rightarrow \ \tan\theta = 2\sqrt{2}\ \ \Rightarrow \ \theta = \tan^{- 1}{(2\sqrt{2})}$