factor and simply sec^4 x - tan^4 x all over (sec^2 x + tan^2 x)
=1
I keep getting cos^2 x + sin^4 x what am I doing wrong?
2007-06-09 13:27:30 · 5 answers · asked by sidd23567 2 in Science & Mathematics ➔ Mathematics
please show me some steps
2007-06-09 13:33:51 · update #1
I realized my mistake
thanks for the answers
2007-06-10 05:52:11 · update #2
sec^4 x - tan^4 x is a difference of two squares
hence it factors as
(sec^2 x - tan ^2 x)(sec^2 x + tan ^2 x)
dividing by (sec^2 x + tan ^2 x)
you are left with
(sec^2 x - tan ^2 x)
but we know that sec^2 x = 1 + tan^2x
hence substituting we get:
1 + tan^2x - tan^2x = 1
2007-06-09 13:38:32 · answer #1 · answered by Anonymous · 1⤊ 0⤋
LHS
= (sec^4 x - tan^4 x) / (sec^2 x + tan^2 x)
= (sec^2 x + tan^2 x)(sec^2 x - tan^2 x) / (sec^2 x + tan^2 x)
= (sec^2 x - tan^2 x)
= 1
= RHS
Since 1 + tan^2 x = sec^2 x
2007-06-09 15:25:20 · answer #2 · answered by Kemmy 6 · 1⤊ 0⤋
Noting that sec²x = tan²x + 1:-
(sec²x - tan²x).(sec²x + tan²x) / (2 tan²x + 1)
(1).(2 tan²x + 1) / (2tan²x + 1)
1
2007-06-09 22:02:17 · answer #3 · answered by Como 7 · 1⤊ 0⤋
sec^4 x - tan^4 x = (sec^2x + tan^2x)(sec^2x -tan ^x)
that over (sec^2x + tan^2x) = sec^2x -tan^2 x and that is an identity.
sec^2x - 1 = Tan^2x
2007-06-09 14:00:55 · answer #4 · answered by davidosterberg1 6 · 1⤊ 0⤋
Given expression
={(sec^2x)^2-(tan^2x)^2/
( sec^2x+tan^2x)
=(sec^2x+tan^2x)(sec^2x-
tan^2x)/(sec^2x+tan^2x)
[factorising the numerator]
=sec^2x-tan^2x
=1
2007-06-09 13:35:47 · answer #5 · answered by alpha 7 · 1⤊ 0⤋