What is the derivative of $y = [\frac{tan x - 1}{sec x}]$ $y = [(tanx - 1) / secx]$ ?

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Answer 1 · 2016-04-12T12:19:08

cos x + sin x

$y = (\frac{tan x - 1}{sec x}) = cos x (tan x - 1) = sin x - cos x$ $y = ((tan x-1)/sec x)= cos x(tan x-1)=sin x-cos x$
So, $y' = cos x + sin x$

What is the derivative of y = [(tanx - 1) / secx]y=[tanx−1secx]y = [(tanx - 1) / secx]?