Trig

Ans not known

@ayush_nath i think it will be pi/2
2<x<2 => 0<x+2<4
$sin^{1}(x+2)+sec^{1}(x+2)=\pi/2$ $sec^{1}(x+2)=cos^{1}(\frac{1}{x+2})=\pi/2sin^{1}(x+2)$
$=cos^{1}(x+2)$ Since cos inverse is oneone
$1/(x+2)=x+2$ So
$x=1$ as$x+2$ >$0$

@harshalpatil thanks