Question

SecA-1= secA(1-cosA)

Answer 1

$\sec A-1=\sec A(1-\cos A)\\\\R=\sec A(1-\cos A)\qquad\text{use distributive property}\\\\R=\sec A-\sec A\cos A\qquad\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\R=\sec A-\dfrac{1}{\cos A}\cdot\cos A\qquad\cos A\ \text{are canceled}\\\\R=\sec A-\dfrac{1}{1}\cdot1\\\\R=\sec A-1\\\\L=\sec A-1\\\\L=R\ :)$

Answer 2

Answer by JKismyhusbandbae:

$\mathrm{Manipulating\:left\:side}\\\sec \left(a\right)-1\\Express\:with\:sin,\:cos\\=-1+\frac{1}{\cos \left(a\right)}\\\mathrm{Simplify}\:-1+\frac{1}{\cos \left(a\right)}:\quad \frac{-\cos \left(a\right)+1}{\cos \left(a\right)}\\\frac{1-\cos \left(a\right)}{\cos \left(a\right)}\\\mathrm{Use\:the\:following\:identity:}\:\frac{1}{\cos \left(x\right)}=\sec \left(x\right)\\=\left(1-\cos \left(a\right)\right)\sec \left(a\right)\\\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}$

$\Rightarrow \mathrm{True}$