Consider the expression, f(x)=(sin−1x)2+(cos−1x)2 Here, sin−1x=a and cos−1x=b then f(x)=a2+b2 =(a+b)2+2ab The value of a and b is (sin−1x+cos−1x)2−2sin−1xcos−1x=
π2
4
−2sin−1x(
π
2
−sin−1x) =
π2
4
−πsin−1x+2(sin−1x)2 For minimum and maximum value, f′(x)=0 π