An algebra problem by Mohammad Khaza

$\begin{aligned} \sec x - \tan x & = \frac 1{\cos x} - \frac {\sin x}{\cos x} = \boxed{\dfrac {1-\sin x}{\cos x}} \end{aligned}$

$\sec~x - \tan~x=\dfrac{1-\sin~x}{\cos~x}$

$\dfrac{1}{\cos~x}-\dfrac{\sin~x}{\cos~x}=\dfrac{1-\sin~x}{\cos~x}$

$\dfrac{1-\sin~x}{\cos~x}=\dfrac{1-\sin~x}{\cos~x}$

$\color{#D61F06}\boxed{\large \therefore \mbox{It is an identity.}}$

B=secx-tanx

=1/cosx - sinx/cosx

= 1-sinx/cosx

 =A/cosx          [A=1-sinx}