# How do you solve tanx + secx = 1?

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Nghi N. Share
May 21, 2015

$\sin \frac{x}{\cos} x + \frac{1}{\cos} x = \frac{\sin x + 1}{\cos} x = 1$

$\sin x + 1 = \cos x$ (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

$\sin x . \cos a - \sin a . \cos x = - \cos a = \frac{- \sqrt{2}}{2}$

$\sin \left(x - \frac{\pi}{4}\right) = - \sin \left(\frac{\pi}{4}\right) = \sin \left(\frac{- \pi}{4}\right)$

a. $x - \frac{\pi}{4} = - \frac{\pi}{4}$ --> x = 0

b. $x - \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5 \pi}{4} \to x = \frac{6 \pi}{4} = \frac{3 \pi}{2}$

Check with equation (1)

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

$x = \frac{3 \pi}{2} \to - 1 + 1 = 0$. OK

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