Prove that? # lim_(h->0)(sec(x+h) - sec x)/h = sec x*tan x #
1 Answer
See belw
Explanation:
We want to prove that
# lim_(h->0)(sec(x+h) - sec x)/h = sec x*tan x #
Hopefully, you can identify this as the limit used in a derivative, and so this is the same as procing that:
# d/dx sec(x) = sec x*tan x #
Let
# L = lim_(h->0)(1/(cos(x+h)) - 1/(cos x))/h #
# :. L = lim_(h->0) (cos x - cos(x+h))/(hcos(x+h)cos x) #
# :. L = lim_(h->0)(cos x - (cos xcos h - sin xsin h))/(hcos(x+h)cos x) #
# :. L = lim_(h->0)(cos x - cos xcos h + sin xsin h)/(hcos(x+h)cos x) #
# :. L = lim_(h->0) (cos x(1 - cos h)+sinxsinh)/(hcos(x+h)cos x) #
# :. L = lim_(h->0) (cos x(1 - cos h))/(hcos(x+h)cos x) + (sin xsinh)/(hcos(x+h)cos x) #
# :. L = lim_(h->0) (1 - cos h)/(hcos(x+h)) + (sin x/cos x)lim_(h->0) (sin h)/(hcos(x+h) #
As
Also,
#:. L = lim_(h->0) (1 - cos h)/(hcos(x+h)) + (sin x/cos x)lim_(h->0) (sin h)/(hcos(x+h) #
# :. L = 0+ (sin x/cos x)1/cosx #
# :. L = tanxsecx # QED