F (x) = cosx √ ((1-sinx) / (1 + SiNx)) + SiNx √ ((1-cosx) / (1 + cosx)) to find monotone intervals and ranges on f (π / 4) and (π / 2, π) F (x) = cosx √ ((1-sinx) / (1 + SiNx)) + SiNx √ ((1-cosx) / (1 + cosx)) 1, find f (π / 4) 2, find the monotone interval and range of function on (π / 2, π)

F (x) = cosx √ ((1-sinx) / (1 + SiNx)) + SiNx √ ((1-cosx) / (1 + cosx)) to find monotone intervals and ranges on f (π / 4) and (π / 2, π) F (x) = cosx √ ((1-sinx) / (1 + SiNx)) + SiNx √ ((1-cosx) / (1 + cosx)) 1, find f (π / 4) 2, find the monotone interval and range of function on (π / 2, π)

f(x)=cosx√((1-sinx)/(1+sinx))+sinx√((1-cosx)/(1+cosx))
=√[(1-sinx)(1-sin^2x)/(1+sinx)]+√[(1-cosx)*(1-cos^2x)/(1+cosx)]
=√[(1-sinx)^2(1+sinx)/(1+sinx)]+√[(1-cosx)^2*(1+cosx)/(1+cosx)]
=√(1-sinx)^2+√(1-cosx)^2
=1-sinx+1-cosx
=2-sinx-cosx
=2-2(√2/2sinx+√2/2cosx)
=2-2sin(x+π/4)
f（π/4）=2-2sin(π/4+π/4)=2-2sinπ/2=2-2=0
The function is a decreasing function at (π / 2, π)
-1

Find the minimum value D of the distance between P and a (m, 0) on the parabola y ^ 2 = 4x + 4

Two cases
Point a (m, 0) is a line perpendicular to x, that is, x = M
1. When there is a focus with the parabola, the minimum distance is the ordinate of the intersection with the parabola when x = M
y^2=4m+4
D = y = qur|4m + 4|: D is equal to the absolute value of (4m + 4) under the root sign
2. When there is no intersection
From y ^ 2 = 4x + 4 when y = 0. X = - 1, we can see that the parabola and X axis intersect at (- 1,0) point
D = | m | - | - 1 |
d=|m|-1

Is f (x) = sinx-x equal to f (x) = cosx-1? How to prove it?

No
Should be
f(x)=sinx-x
Then the derivative f '(x) = cosx-1