prove: sec(2theta)= (cos(theta)/(cos(theta)+sin(theta)) + sin(theta)/(cos(theta)-sin(theta))

Answer 1

We can use the double-angle formula for cosine: cos(2Θ) = cos²Θ - sin²Θ, so sec(2Θ) = 1 / (cos²Θ - sin²Θ).

We'll manipulate the right-hand side of the given identity until it is the expression above for sec(2Θ):

cosΘ / [cosΘ + sinΘ] + sinΘ / [cosΘ - sinΘ] Multiply each fraction top and bottom by the denom. of other:

= cosΘ [cosΘ - sinΘ] / [cosΘ - sinΘ][cosΘ + sinΘ] + sinΘ[cosΘ + sinΘ] / [cosΘ + sinΘ][cosΘ - sinΘ]

= [cos²Θ - sinΘcosΘ + sinΘcosΘ + sin²Θ] / (cos²Θ - sin²Θ)

= [cos²Θ + sin²Θ] / (cos²Θ - sin²Θ)

= 1 / (cos²Θ - sin²Θ)

= 1 / cos(2Θ)

= sec2Θ ◊