Question

How do you solve `tanx + secx = 1`?

Answer 1

`sin x/cos x + 1/cos x = (sin x + 1)/cos x = 1`

`sin x + 1 = cos x` (1)
sin x - cos x = -1. Call a whose tan is tan a = 1 = tan pi/4
sin x -(sin a/cos a) sin x = -1

`sinx.cos a - sin a.cosx = -cos a = (-sqrt2)/2`

`sin(x - (pi)/4) = -sin (pi/4) = sin ((-pi)/4)`

a. `x - (pi)/4 = -pi/4` --> x = 0

b. `x - pi/4 = pi + pi/4 = (5pi)/4 -> x = (6pi)/4 = (3pi)/2`

Check with equation (1)

x = 0 --> sin x + 1 = cos x --> 0 + 1 = 1 . OK

`x = (3pi)/2 -> -1 + 1 = 0`. OK