Question

tan（π/2+α）=-cosα

Answer 1

(tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α)

= (tan(π/4+α)·cos2α)/2sin(π/2-(π/4-α))

= (tan(π/4+α)·cos2α)/2sin(π/4+α)

=cos2α/(2sin(π/4+α))

=sin(π/2-2α)/(2sin(π/4+α))

= 2sin(π/4+α) cos(π/4+α) / (2sin(π/4+α))

= cos(π/4+α)

Answer 2

Answer 3

(tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α)

= (tan(π/4+α)·cos2α)/2sin(π/2-(π/4-α))

= (tan(π/4+α)·cos2α)/2sin(π/4+α)

=cos2α/(2sin(π/4+α))

=sin(π/2-2α)/(2sin(π/4+α))

= 2sin(π/4+α) cos(π/4+α) / (2sin(π/4+α))

= cos(π/4+α)

Answer 4

(tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α)

= (tan(π/4+α)·cos2α)/2sin(π/2-(π/4-α))

= (tan(π/4+α)·cos2α)/2sin(π/4+α)

=cos2α/(2sin(π/4+α))

=sin(π/2-2α)/(2sin(π/4+α))

= 2sin(π/4+α) cos(π/4+α) / (2sin(π/4+α))

= cos(π/4+α)

Answer 5

(tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α) (tan(π/4+α)·cos2α)/2cos(π/4-α)

= (tan(π/4+α)·cos2α)/2sin(π/2-(π/4-α))

= (tan(π/4+α)·cos2α)/2sin(π/4+α)

=cos2α/(2sin(π/4+α))

=sin(π/2-2α)/(2sin(π/4+α))

= 2sin(π/4+α) cos(π/4+α) / (2sin(π/4+α))

= cos(π/4+α)