How do you simplify #(sec^2x-1)/(secx-1)#?
1 Answer
Mar 9, 2017
Explanation:
The numerator is a
#color(blue)"difference of squares"# and is factorised, in general, as shown.
#color(red)(bar(ul(|color(white)(2/2)color(black)(a^2-b^2=(a-b)(a+b))color(white)(2/2)|)))#
#"here "a=secx" and "b=1#
#rArrsec^2x-1=(secx-1)(secx+1)#
#rArr(sec^2x-1)/(secx-1)#
#=(cancel((secx-1)^1)(secx+1))/cancel(secx-1)^1#
#=secx+1#
