Hey guys can you verify my logic?

If asec(x)=1+tan(x)

and bsec(x)=1-tan (x)

prove a^2 + b^2 = 2

a = 1+tan(x) / sec(x)

a = 1+tan(x) * cos(x)

a = cos + sin

b = 1-tan(x) / sec(x)

b = 1-tan(x) * cos(x)

b = cos(x) - sin(x)

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

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

= 2( cos^2(x) + sin^2(x) )

= 2(1)

= 2