How do find the derivative of $y = \frac{1 - sec x}{tan x}$ $y= (1- sec x)/ tan x$ ?

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Answer 1 · 2018-07-30T18:53:04

$\frac{d y}{d x} = csc x (cot x - csc x)$ $(dy)/(dx)=cscx(cotx-cscx)$

Here ,

$y = \frac{1 - sec x}{tan x}$ $y=(1-secx)/tanx$

$\Rightarrow y = \frac{1}{tan x} - \frac{sec x}{tan x}$ $=>y=1/tanx-secx/tanx$

$\Rightarrow y = cot x - \frac{\frac{1}{cos x}}{\frac{sin x}{cos x}}$ $=>y=cotx-(1/cosx)/(sinx/cosx)$

$\Rightarrow y = cot x - csc x$ $=>y=cotx-cscx$

$\frac{d y}{d x} = - {csc}^{2} x - (- csc x cot x)$ $(dy)/(dx)=-csc^2x-(-cscxcotx)$

$\frac{d y}{d x} = - {csc}^{2} x + csc x cot x$ $(dy)/(dx)=-csc^2x+cscxcotx$

$:.(dy)/(dx)=cscx(cotx-cscx)$

How do find the derivative of y= (1- sec x)/ tan xy=1−secxtanxy= (1- sec x)/ tan x?