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Q.
Sum of the squares of all the solution(s) of the equation, $2 \sin ^{-1}(x+2)=\cos ^{-1}(x+3)$ is
Inverse Trigonometric Functions
Solution:
$\text { Let } \sin ^{-1}( x +2)=\alpha x +2=\sin \alpha$
$\therefore 2 \alpha=\cos ^{-1}( x +3) $
$\cos 2 \alpha= x +3=( x +2)+1=1+\sin \alpha $
$1-2 \sin ^2 \alpha=1+\sin \alpha $
$\sin \alpha(1+2 \sin \alpha)=0$
$\Rightarrow \sin \alpha=0 \text { or } \sin \alpha=-1 / 2 $
$\therefore x =-2 \text { or } x =-2.5 \text { (rejected) as it does not satisfy the original equation } $
$\therefore \therefore x ^2=6.25$