# Trig

• Ans not known

• @ayush_nath i think it will be pi/2

-2<x<2 => 0<x+2<4

$sin^{-1}(x+2)+sec^{-1}(x+2)=\pi/2$

$sec^{-1}(x+2)=cos^{-1}(\frac{1}{x+2})=\pi/2-sin^{-1}(x+2)$
$=cos^{-1}(x+2)$

Since cos inverse is one-one

$1/(x+2)=x+2$

So $x=-1$ as $x+2$ > $0$

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