prove that 1/(sec x -tan x)-1/cos x = 1/cos x - 1/(sec x + tan x )

Asked by  | 10th Aug, 2012, 05:47: PM

Expert Answer:

LHS = 1/(sec x -tan x)-1/cos x
= sec x + tan x/(sec x - tan x)(sec x + tan x) - 1/cos x
= sec x + tan x/(sec2x - tan2x) - 1/cos x
= sec x + tan x - sec x
= tan x
 
RHS = 1/cos x - 1/(sec x + tan x)
1/cos x - (sec x - tan x)/(sec x + tan x) (sec x - tan x)
= 1/cos x - (sec x - tan x)/(sec2x - tan2x)
= sec x - (sec x - tan x)/1
= sec x - sec x + tan x
= tan x
 
Hence, LHS = RHS

Answered by  | 10th Aug, 2012, 09:31: PM

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