i think i have it setup correctly but im not sure

Question

Solve the following equation for θ where 0≤θ<π.

sec(−θ+π/2)=csc(−θ+π/2)

what i have is

(−θ+π/2)+(−θ+π/2)

Answer 1

Hi Dylan,

Let x= -θ+π/2

Then sec(x)=csc(x)

Multiply both sided by sin(x)

sin(x)/cos(x)=1 or

tan (x)=1

so x=tan^-1(1) =π/4+nπ, n=0,1,2.....

but x=-θ+π/2

so -θ+π/2=π/4+nπ,

solve for θ.

Jim

Answer 2

Let a = theta if sec(-a+pi/2) = csc(-a+pi/2) then cos(-a+pi/2) = sin(-a+pi/2)

Using the addition identities we have cos(-a)cos(pi/2)-sin(-a)sin(pi/2) = sin(-a)cos(pi/2)+cos(-a)sin(pi/2)

Then

0 + sin(a) = 0 + cos(a)

Or Sin(a)/cos(a) = 1 Tan(a) Between 0 & pi a = 45 degrees (pi/4)

Answer 3

sec (π/2 - θ) = 1/cos (π/2 - θ)

1/cos (π/2 - θ) = 1/ sin θ

csc (π/2 - θ) = 1/sin (π/2 - θ)

1/sin (π/2 - θ) = 1 / cos θ

1 / sin θ = 1 / cos θ

sin θ = cos θ

θ = π/4 + kπ