Trig

Ans not known

@ayush_nath i think it will be pi/2
2<x<2 => 0<x+2<4
$sin_{−1}(x+2)+sec_{−1}(x+2)=π/2$
$sec_{−1}(x+2)=cos_{−1}(x+21 )=π/2−sin_{−1}(x+2)$
$=cos_{−1}(x+2)$Since cos inverse is oneone
$1/(x+2)=x+2$
So $x=−1$ as $x+2$ > $0$

@harshalpatil thanks