X is (0, π / 2). If sinxcosx = 1 / 2, find the value of 1 / (1 + SiNx) + 1 / (1 + cosx)

From sin (x) cos (x) = 1 / 2
(sin(x)+cos(x))^2
=sin^2(x)+cos^2(x)+2sin(x)cos(x)
=1+2*1/2
=2,
So sin (x) + cos (x) = sqrt (2) (from X in the first quadrant, sin (x) > 0, cos (x) > 0),
therefore
1/(1+sin(x)) +1/(1+cos(x))
=[(1+sin(x))+(1+cos(x))]/[(1+sin(x)) (1+cos(x))]
=[2+sin(x)+cos(x)]/[1+sin(x)+cos(x)+six(x)cos(x)]
=(2+Sqrt(2))/(1+Sqrt(2)+1/2)
=(4+2*Sqrt(2))/(3+2*Sqrt(2))
=(4+2*Sqrt(2))(3-2*Sqrt(2))
=4-2*Sqrt(2)

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