Given that y= x- ln (sec x+tan x), 0 <=x<= pi/2.

By finding dy/dx or otherwise, deduce that y<0 for 0<x<pi/2.

Although I can find the answer for dy/dx, which is 1- sec x (i try to express it in one trigonometic function), but i have the diifficulties in deducing that y<0 for 0<x<pi/2. Can anyone help me?

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You need to check if derivative of the given function is negative if `x in (0, pi/2)` to prove that the function negative for `x in (0,pi/2)` such that:

`(dy)/(dx) = 1 - ((sec x + tan x)')/(sec x + tan x)`

`(dy)/(dx) = 1 - (sin x/(cos^2 x) + 1/(cos^2 x))/((1 + sin x)/cos x)`

`(dy)/(dx) = 1 - ((1 + sin x)/(cos^2 x))/((1 + sin x)/cos x)`

Reducing duplicate factors yields:

`(dy)/(dx) = 1 - 1/cos x = 1 - sec x`

You need to consider the following value for x such that:

`x = pi/4 => (dy)/(dx)|_(x=pi/4) = 1 - 1/(cos (pi/4))`

`(dy)/(dx)|_(x=pi/4) = 1 - 2/sqrt2 => (dy)/(dx)|_(x=pi/4) = 1 - 2sqrt2/2`

`(dy)/(dx)|_(x=pi/4) = 1 - sqrt2`

Notice that `1 - sqrt 2 < 0 => (dy)/(dx)|_(x=pi/4) < 0` , hence, `(dy)/(dx) < 0` if `x in (0,pi/2).`

You need to remember that if deriative of function is negative, then the function decreases, thus is negative.

**Hence, since `(dy)/(dx) < 0` if `x in (0,pi/2)` , then `y < 0` , if `x in (0,pi/2).` **

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