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August 1, 2015

August 1, 2015

Posted by **sumayya** on Monday, June 29, 2009 at 2:31pm.

_______1______ - ___1___ =__1___

sec x-tan x cos x cos x

- _______1______

sec x + tan x

- maths -
**MathMate**, Monday, June 29, 2009 at 6:32pmI will rewrite the question as:

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

transpose terms to get:

1/(sec(x)-tan(x)) + 1/(sec(x)+tan(x)) = 2/cos(x)

Add the two fractions on the left and simplify using sec^{2}(x)=1+tan^{2}(x) on the left-hand side to get 2sec(x)=2/cos(x).