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

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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