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=(1)/(\cos x)\\\\R=\sec A-(1)/(\cos A)\cdot\cos A\qquad\cos A\ \text{are canceled}\\\\R=\sec A-(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+(1)/(\cos \left(a\right))\\\mathrm{Simplify}\:-1+(1)/(\cos \left(a\right)):\quad (-\cos \left(a\right)+1)/(\cos \left(a\right))\\(1-\cos \left(a\right))/(\cos \left(a\right))\\\mathrm{Use\:the\:following\:identity:}\:(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}`