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

Question

31760 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-21T17:46:41

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

$sin x + 1 = cos x$ (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

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