Match the following

Column - I		Column - II
A)	The maximum value of  $\sin \left(\cos x\right)+\cos \left(\sin x\right),x\in [-\frac{\pi }{2},\frac{\pi }{2}]$ is	p)	$\cos \left(\cos 1\right)$
B)	The minimum value of$\sin \left(\cos x\right)+\cos \left(\sin x\right),x\in [-\frac{\pi }{2},\frac{\pi }{2}]$  is	q)	$1+\cos 1$
C)	The maximum value of $\cos \left(\cos \left(\sin x\right)\right)$ is	r)	$\cos 1$
D)	The minimum value of $\cos \left(\cos \left(\sin x\right)\right)$ is	s)	$1+\sin 1$

Let  $f\left(x\right)=\sin \left(\cos x\right)+\cos \left(\sin x\right)$
$f$  is an even function. We can take $x\in [0,\frac{\pi }{2}].In[0,\frac{\pi }{2}],\sin x$  increasing and $\cos x$  is decreasing.
Hence  $f$ is decreasing function. Therefore, maximum value of $f$  is  $f\left(0\right)=\sin 1+1$ and minimum value is  $f\left(\frac{\pi }{2}\right)=0+\cos 1.$
Let  $g\left(x\right)=\cos \left(\cos \left(\sin x\right)\right).$ Obviously   is an even periodic function of period  $\pi .$
Hence $g$  takes all of its values for  $x\in [0,\frac{\pi }{2}].$
It can be seen that $g$  is an increasing function in $[0,\frac{\pi }{2}].$  So maximum value of  $g=g\left(\frac{\pi }{2}\right)=\cos \left(\cos 1\right),$ and minimum value of  $g=g\left(0\right)=\cos 1.$