Question #0038f

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Answer 1 · 2017-02-18T08:26:59

see 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)