Expert: Sherman D. - 1/13/2009

Question
((2tanx)/(1-tan^2x))+((1)/(2cos^2x-1))=(cosx+sinx)/(cosx-sinx)

Answer
((2tan(x))/(1 - tan(x)^2))) + (1/(2cos(x)^2 - 1))) =
(cos(x) + sin(x))/(cos(x) - sin(x))

((2tan(x))/(1 - tan(x)^2))) + (1/(2cos(x)^2 - 1)))

i'll do this the long way for you.

(2(sin(x)/cos(x))/(1 - (sin(x)/cos(x))^2)) + (1/(2cos(x)^2 - 1))) =

(2sin(x)/cos(x))/((cos(x)^2 - sin(x)^2)/cos(x)^2))) + (1/(2cos(x)^2 - 1)) =

(2sin(x)cos(x)^2/(cos(x)(cos(x)^2 - sin(x)^2))) + (1/(2cos(x)^2 - 1)) =

(2sin(x)cos(x))/(cos(x)^2 - sin(x)^2) + (1/(2cos(x)^2 - 1)) =

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

now for the right side of the equation.

(cos(x) + sin(x))/(cos(x) - sin(x))

multiply top and bottom by cos(x) + sin(x)

(cos(x)^2 + 2sin(x)cos(x) + sin(x)^2)/(cos(x)^2 - sin(x)^2)

(2sin(x)cos(x) + (cos(x)^2 + sin(x)^2))/(cos(x)^2 - sin(x)^2)

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

More info found at
http://math2.org/math/trig/identities.htm