# prove: sec x + tan x = 1/(sec x - tan x) IN LHS = RHS FORM?

also: sec x + (cosec x)(cot x) = (sec x)(cosec^2 x) also in lhs=rhs form

cheers

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### 4 Answers

- DambarudharLv 78 years agoFavorite Answer
(1) LHS = (sec x + tan x) = (sec x + tan x) / 1

= (sec x + tan x) / (sec² x - tan² x)

= (sec x + tan x) / {(sec x + tan x)(sec x - tan x)}

= 1 / (sec x - tanx) = RHS

(2) LHS = sec x + (cosec x)(cot x) = (1 / cos x) + {(cos x) / (sin² x)}

= (sin² x + cos² x) / { (cos x)(sin² x) }

= 1 / { (cos x)(sin² x) }

= (1 / cos x)*(1 / sin² x) = (sec x)(cosec² x) = RHS

- 8 years ago
sec x = 1/cosx and tanx = sin x/cos x, so adding the LHS, you have (1+sin x)/cosx

On the RHS, you have 1/ (1/cosx - sinx/cosx) which is 1/(1-sinx)/cosx = cos x /(1-sinx).

Your equation now is (1+sin x)/cosx = cos x /(1-sinx). Cross multiply and you get: cos^2 x = 1 - sin^2 x

Which is true by the Pythagorean Identity,

Source(s): B.S in mathematics - ambachLv 43 years ago
once you go-multiply you've (secx+tanx)(secx-tanx)=a million huge difference of squares so sec^2x-tan^2x=a million a million/cos^2x-sin^2x/cos^2x= (a million-sin^2x)/cos^2x= cos^2x/cos^2x=a million it really is what we would have beloved!