Question

Question #0038f

Answer 1

see explanation

Explanation:

`1/(sec x - tanx) + 1/(secx+tanx) = 2/cosx`

equate the denominator

`(sec x+tan x + sec x - tan x)/((sec x - tanx) *(secx+tanx)) = 2/cosx`

`(2sec x) /(sec ^2x - tan ^2x) = 2/cos x`

make sec and tan become fraction

`(2sec x )/ (1/cos^2 x - sin^2 x/cos ^2 x)= 2/cosx`

`(2sec x )/((1 - sin^2 x)/cos ^2 x)= 2/cosx`

use trig identity so `1-sin^2x =cos ^ 2 x`

`(2sec x )/((cos^2x)/cos ^2 x)= 2/cosx`

`(2sec x )/ 1= 2/cosx`

`2 * 1/cos x = 2/cos x`

`2/cosx = 2/cos x` (verified)