How do you solve $ \sec x - 1 = \tan x? $

Hint:Try to convert the equation in form of $ \sin \;\& \;\cos $ and then solve further in order to make the equation parallel to the compound angle formula of trigonometric identities you should divide the converted equation of $ \sin \;{\text{and}}\;\cos $ with $ \sqrt {{{(coefficient\;of\;\sin x)}^2} + {{(coefficient\;of\;\cos x)}^2}} $ in order to get the required equation in which you can easily apply compound angle formulas of trigonometric identities.

Complete step by step solution:
Given $ \sec x - 1 = \tan x $ , we have to convert it into $ \sin
\;{\text{and}}\;\cos $ form, to do this we will divide both sides with $ \sec x $
 \[
\Rightarrow \sec x - 1 = \tan x \\
\Rightarrow \dfrac{{\sec x - 1}}{{\sec x}} = \dfrac{{\tan x}}{{\sec x}} \\
\Rightarrow \dfrac{{\sec x}}{{\sec x}} - \dfrac{1}{{\sec x}} = \dfrac{{\tan x}}{{\sec x}} \\
\Rightarrow 1 - \dfrac{1}{{\sec x}} = \dfrac{{\tan x}}{{\sec x}} \\
\]

Now we know that $ \sec x = \dfrac{1}{{\cos x}}\;\& \;\tan x = \dfrac{{\sin x}}{{\cos x}} $ , so replacing
them with $ \sin \;{\text{and}}\;\cos $ as
 \[
\Rightarrow 1 - \dfrac{1}{{\dfrac{1}{{\cos x}}}} = \dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\cos
x}}}} \\
\Rightarrow 1 - \cos x = \sin x \\
\Rightarrow \cos x + \sin x = 1 \\
\]

Now, in order to make this equation comparable to compound trigonometric formula, we will divide it
by $ \sqrt {{{(coefficient\;of\;\sin x)}^2} + {{(coefficient\;of\;\cos x)}^2}} $
 \[
\Rightarrow \cos x + \sin x = 1 \\
\Rightarrow \dfrac{{\cos x + \sin x}}{{\sqrt {{1^2} + {1^2}} }} = \dfrac{1}{{\sqrt {{1^2} + {1^2}} }}
\\
\Rightarrow \dfrac{{\cos x + \sin x}}{{\sqrt {1 + 1} }} = \dfrac{1}{{\sqrt {1 + 1} }} \\

\Rightarrow \dfrac{{\cos x + \sin x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\
\Rightarrow \dfrac{{\cos x}}{{\sqrt 2 }} + \dfrac{{\sin x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\
\]

Now we all know that \[\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\] , so replacing
 $ \dfrac{1}{{\sqrt 2 }} $ with $ \cos \dfrac{\pi }{4}\;{\text{and}}\;\sin \dfrac{\pi }{4} $ , we will get
 $ \Rightarrow \cos x\cos \dfrac{\pi }{4} + \sin x\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} $

We have seen this type of equation before, do you remember where?
We have seen this type of trigonometric equation before in the compound angle formula of
trigonometric identities. Now, this becomes a trigonometric identity which is similar to this
 $ \cos x\cos y + \sin x\sin y = \cos (x - y) $

Now using the cosine formula of compound angle to solve further, we can write it as
 $
\Rightarrow \cos x\cos \dfrac{\pi }{4} + \sin x\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \\
\Rightarrow \cos (x - \dfrac{\pi }{4}) = \dfrac{1}{{\sqrt 2 }} \\
 $

Now we know the general solution of $ \cos \theta = \dfrac{1}{{\sqrt 2 }} $ , which is $ x = 2n\pi \pm
\dfrac{\pi }{4},\;{\text{where}}\;n \in I $
 \[
\Rightarrow \cos \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\
\Rightarrow x - \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4},\;{\text{where}}\;n \in I \\
\Rightarrow x = 2n\pi \pm \dfrac{\pi }{4} + \dfrac{\pi }{4},\;{\text{where}}\;n \in I \\
\]

Checking for $ x = 2n\pi + \dfrac{\pi }{4} + \dfrac{\pi }{4} = 2n\pi + \dfrac{\pi }{2} $ here $ \sec x $ and
 $ \tan x $ are undefined.

Checking for $ x = 2n\pi - \dfrac{\pi }{4} + \dfrac{\pi }{4} = 2n\pi $ , $ \sec x = 1 $ and $ \tan x = 0 $ which
is satisfying $ \sec x - 1 = \tan x $

$ \therefore $ required solution is $ x = 2n\pi $

Note: We can solve it by one more method,
Given $ \sec x - 1 = \tan x $
 $ \Rightarrow \sec x - \tan x = 1 $ _____(I)
We know that $ {\sec ^2}x - {\tan ^2}x = 1 $
 $
\Rightarrow {\sec ^2}x - {\tan ^2}x = 1 \\
\Rightarrow (\sec x + \tan x)(\sec x - \tan x) = 1 \\
 $
Now using equation (I) and substituting the value of $ \sec x - \tan x = 1 $ in above equation in order to solve further
 \[ \Rightarrow (\sec x + \tan x) \times 1 = 1\]
 $ \Rightarrow \sec x + \tan x = 1 $ ______(II)

Now adding equation (I) and (II) we will get,
 $
\Rightarrow \sec x - \tan x + \sec x + \tan x = 1 + 1 \\
\Rightarrow 2\sec x = 2 \\
\Rightarrow \sec x = \dfrac{2}{2} \\
\Rightarrow \sec x = 1 \\
 $

Now we know that $ \sec x = \dfrac{1}{{\cos x}}\; $
 $
\Rightarrow \sec x = 1 \\
\Rightarrow \dfrac{1}{{\cos x}} = 1 \\
\Rightarrow 1 = \cos x \\
 $

Now we know the general solution for $ \cos x = 1 $ is $ 2n\pi $ , where \[n \in I\]
 $ \therefore $ required solution is $ x = 2n\pi $