2007-01-30 15:26:37 · 8 answers · asked by Richard H 2 in Science & Mathematics ➔ Mathematics
No, to see that this is not true, you could just plug in 0. However it is true that 1+ (tanx)^2 = (secx)^2.
Fortunately, you do not have to remember this -- I sure as heck can't. You just have to remember that:
(sinx)^2 + (cosx)^2 = 1
Then to get a relation with tanx and secx, just divide all the terms by (cosx)^2 to get:
(tanx)^2 + 1 = (secx)^2
You can also divide (sinx)^2 + (cosx)^2 = 1 through by (sinx)^2 to get:
1 + (cotx)^2 = (cscx)^2
2007-01-30 15:29:39 · answer #1 · answered by Phineas Bogg 6 · 0⤊ 0⤋
No - that's backwards. We know sin^2 x + cos^2 x = 1. Divide through by cos^2 x, and we get tan^2 x + 1 = sec^2 x, which is the correct identity. tan^2 x = sec^2 x - 1.
2007-01-30 15:32:07 · answer #2 · answered by airtime 3 · 0⤊ 1⤋
1 + sec(x)^2 = 1 + (1/cos(x))^2 = (1 + cos(x)^2)/(cos(x)^2))
tan(x)^2 = (sin(x)/cos(x))^2 = (1 - cos(x)^2)/(cos(x)^2)
so for 1 + sec(x)^2 to equal tan(x)^2, you have to have 1 - sec(x)^2
2007-01-30 15:48:14 · answer #3 · answered by Sherman81 6 · 1⤊ 0⤋
yes
2007-01-30 15:29:47 · answer #4 · answered by Anonymous · 0⤊ 3⤋
yes
2007-01-30 15:28:45 · answer #5 · answered by Lala Girl 2 · 0⤊ 3⤋
no
2007-01-30 15:32:50 · answer #6 · answered by m m 3 · 1⤊ 0⤋
www.calculator.com
2007-01-30 15:29:53 · answer #7 · answered by vlad|KARNAFËL 【KOR】 3 · 0⤊ 1⤋
Indeed it is.
2007-01-30 15:28:50 · answer #8 · answered by Steady As She Goes 2 · 0⤊ 3⤋