# Solve the equation secx=1+tanx?

Jan 25, 2017

$x = 2 n \pi$ or $x = 2 n \pi + \frac{\pi}{2}$

#### Explanation:

$\sec x = 1 + \tan x$ can be written as

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

and assuming $\cos x \ne 0$ this

$1 = \cos x + \sin x$

or $\sqrt{2} \left(\sin x \cos \left(\frac{\pi}{4}\right) + \cos x \sin \left(\frac{\pi}{4}\right)\right) = 1$

or $\sqrt{2} \sin \left(x + \frac{\pi}{4}\right) = 1$

or $\sin \left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}\right)$

Hence $x + \frac{\pi}{4} = n \pi + {\left(- 1\right)}^{n} \left(\frac{\pi}{4}\right)$

or $x = n \pi + {\left(- 1\right)}^{n} \left(\frac{\pi}{4}\right) - \frac{\pi}{4}$

which simplifies to $x = 2 n \pi$ or $x = 2 n \pi + \frac{\pi}{2}$

Jan 25, 2017

Equation given

$\sec x = 1 + \tan x$

$\implies \sec x - \tan x = 1. \ldots . \left(1\right)$

Again we know

${\sec}^{2} x - {\tan}^{2} x = 1. \ldots \left(2\right)$

Dividing (2) by (1) we get

$\sec x + \tan x = 1. \ldots \left(3\right)$

Adding (1) and (3) we get

$2 \sec x = 2$

$\implies \cos x = 1$

$\implies x = 2 n \pi \text{ where } n \in \mathbb{Z}$

Putting $\cos x = 1$ in equation (1) we get
$\tan x = 0$

$\implies x = n \pi \text{ where } n \in \mathbb{Z}$

but for n odd $\cos x = \cos n \pi = - 1$

So the only solution is $x = 2 n \pi$

Jun 12, 2018

$x = 2 k \pi$

#### Explanation:

$\frac{1}{\cos x} = 1 + \sin \frac{x}{\cos x}$
$1 = \sin x + \cos x$ (condition $\cos x \ne 0$)
$\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 1$
$\cos \left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Trig table and unit circle give 2 solutions for $\left(x - \frac{\pi}{4}\right)$-->
$x - \frac{\pi}{4} = \pm \frac{\pi}{4}$
a. $x - \frac{\pi}{4} = \frac{\pi}{4} + 2 k \pi$
$x = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} + 2 k \pi$
This answer is rejected because of the above condition:
(cos x != 0 --> x != pi/2, and x != (3pi)/2)
b. $x - \frac{\pi}{4} = - \frac{\pi}{4} + 2 k \pi$
$x = 2 k \pi$
Check.
$x = 2 \pi$--< cos x = 1 --> sec x = 1/1 = 1 --> tan x = 0
1 + tan x = 1 + 0 = 1. Proved.