Prove each of the following identities algebraically:
a) cos(x) + (cos(x)*tan(x)^2 = sec(x)
b) cos(x)(sec(x)-csc(x))=1-cot(x)
cos(x) + cos(x) tan(x)^2 = cos(x) + cos(x) sin^2(x)/cos^2(x)
................. = cos(x) + sin^2(x)/cos(x) = [cos^2(x) + sin^2(x)]/cos(x)
................. = 1/cos(x) = sec(x)
cos(x)(sec(x)-csc(x)) = cos(x) [1/cos(x) - 1/sin(x)]b) cos(x)(sec(x)-csc(x)) = 1-cot(x)
.................... = cos(x) [ (sin(x) - cos(x))/(cos(x) sin(x) ]
.................... = (sin(x) - cos(x))/ sin(x)
.................... = 1 - cos(x)/sin(x) = 1- cot(x)
RonL
Hello, brewerangel!
Another approach . . .
The left side is: .cos(x) + cos(x)·tan²(x)a) .cos(x) + cos(x)·tan²(x) .= .sec(x)
Factor: . . . . . . .cos(x)·[1 + tan²(x)]
Then we have: . cos(x)·sec²(x) .= .cos(x)·sec(x)·sec(x)
Since cos(x)·sec(x) = 1, we have: .sec(x)
The left side is: .cos(x)·sec(x) - cos(x)·csc(x)b) .cos(x)[sec(x)-csc(x)] .= .1 - cot(x)
. . . . . . . . . . . . . . cos(x)
And we have: . .1 - -------- . = . 1 - cot(x)
. . . . . . . . . . . . . . sin(x)