2007-03-09 15:55:16 · 6 answers · asked by saum_leo 1 in Science & Mathematics ➔ Mathematics
1-sin^2 theta /1-cos ^2theta
=cos^2 theta/sin^2 theta
=cot^2 theta..................
2007-03-09 18:01:36 · answer #1 · answered by Anonymous · 1⤊ 0⤋
1 - sin^2 theta = cos^2 theta
1 - cos^ theta = sin^2 theta
cos^2 theta/sin^2 theta = cot^2 theta.
2007-03-09 15:59:22 · answer #2 · answered by Newbody 4 · 0⤊ 0⤋
1-sin square theta=cos square theta
1- cos square theta=sin square theta
cos square theta/sin square theta =cot square theta
2007-03-09 21:37:32 · answer #3 · answered by Aneeqa 4 · 0⤊ 0⤋
cot square theta
2007-03-09 20:41:44 · answer #4 · answered by Somebody 2 · 0⤊ 0⤋
Let theta = Ø (because I can`t find theta!)
[1 - sin ² Ø ] / [ 1 - cos ² Ø ]
= cos ² Ø / sin ² Ø
= cot ² Ø
2007-03-10 00:48:24 · answer #5 · answered by Como 7 · 0⤊ 0⤋
try to keep all the formulas of trignometry in your mind and substitute it in place of them and take theta as A
so we know that sin^2A+cos^A=1
so [1-sin^2A]/[1-cos^2A]]
= cos^2A/sin^2A
=cot^2A
2007-03-10 19:26:26 · answer #6 · answered by suchi 2 · 0⤊ 0⤋