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

1 Answer
Nghi N.
May 21, 2015

sin x/cos x + 1/cos x = (sin x + 1)/cos x = 1sinxcosx+1cosx=sinx+1cosx=1

sin x + 1 = cos xsinx+1=cosx (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)/2sinx.cosa−sina.cosx=−cosa=−√22

sin(x - (pi)/4) = -sin (pi/4) = sin ((-pi)/4)sin(x−π4)=−sin(π4)=sin(−π4)

a. x - (pi)/4 = -pi/4x−π4=−π4 --> x = 0

b. x - pi/4 = pi + pi/4 = (5pi)/4 -> x = (6pi)/4 = (3pi)/2x−π4=π+π4=5π4→x=6π4=3π2

Check with equation (1)

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

x = (3pi)/2 -> -1 + 1 = 0 x=3π2→−1+1=0. OK

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