y= tan^-1(secx-tanx), then dy/dx

d (tan-1(x))/dx = 1/(1+x^2)
To prove the above result,
let y=tan-1(x)
or,tan(y)=x, differentiate w.r.t. x,
sec^2(y).dy/dx=1, or,
dy/dx=cos^2(y) ...(i)
Let p=tan^-1(x), or x=tan(p), so cos(p)=1/(1+x^2)^0.5,
So, cos(p)=cos(tan-1(x))=1/(1+x^2)^0.5.
Therefore, dy/dx=cos^2(y)=cos^2(tan-1(x))=1/(1+x^2)^0.5

In the original question
dy/dx=(sec(x)tan(x)-sec^2(x)) / (1 + (sec(x) - tan(x))^2)
       =-1/2 (after simplification, using 1+tan^2(x)=sec^2(x))

derivative of arc tan x is 1/(1+x²)

here dy/dx= ( secx.tanx - sec²x) / (1+sec²x+tan²x-2.secx.tanx)

               = secx(secx-tanx)/2.tanx(secx-tanx)

               =secx/2tanx

               =(cosec x) /2

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regards

Ramesh