Trigonometry question

derivative of   y=1-secx/tanx            We have    sec x  = 1/cos x and tan x = sin x / cos x

so  sec x/tan x    =  (1/cosx)/(tanx) = (1/cosx)   X   ( cosx/sinx)   =  1/sinx

then y= 1-1/sinx

dy/dx  = 0  - d/dx(-1/sinx)   = d/dx(  (sinx)^(-1) ).  Now use Chain Rule

d/dx ( (sinx)^(-1) )    = (-sinx)^(-2)   X     cosx

= (-cosx)/(sinx)(sinx)= -tanx cosecx.

$y = 1-\dfrac{\sec x}{\tan x} =1 - \sec x \cot x = 1- \dfrac{1}{\cos x \tan x}=1-\dfrac{1}{\sin x}\\ y'=(1-\dfrac{1}{\sin x})'=0-(\dfrac{\cos x}{\sin^2x})=-\dfrac{\cos x}{\sin^2x}$