Help please, proving trig identities
tan^2@ - sin^2@ = sin^4@sec^2@
(@ = theta)
2007-02-27 20:53:38 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Will use Ø as I can`t find theta!
LHS
sin ² Ø / cos ² Ø - sin ² Ø
= [sin ² Ø - sin ² Ø.cos ² Ø] / cos ² Ø
= sin ² Ø. [1 - cos ² Ø] / cos ² Ø.
= sin ² Ø.[ sin ² Ø ] / cos ² Ø
= sin^(4) Ø. sec ² Ø = RHS
2007-02-27 21:41:51 · answer #1 · answered by Como 7 · 0⤊ 0⤋
ok the answer is really simple:
we use the following formulas:
tg^2(t)=sin^2(t)/cos^2(t)
1-cos^2(t)=sin^2(t)
1/cos^2(t)=sin^2(t)
ok lets go to ur question:
tg^2(t)-sin^2(t)=
sin^2(t)/cos^2(t)-sin^2(t)=
sin^2(t)(1/cos^2(t)-1)=
sin^2(t)((1-cos^2(t))/cos^2(t))=
sin^2(t)(sin^2(t))/cos^2(t)=
sin^4(t)*sec^2(t) and the proof
is complete.
2007-02-27 21:44:16 · answer #2 · answered by vinchenzo_corleone 2 · 0⤊ 0⤋
tan^2@-sin^2@=
sin^2@/cos^2@-sin^2@=
(sin^2@-sin^2@cos^2@)/cos^2@=
sin^2@(1-cos^2@)/cos^2@=
***note***sin^2@+cos^2@=1**
sin^2@(sin^2@)/cos^2@=
sin^4@sec^2@=
That help?
2007-02-27 21:36:48 · answer #3 · answered by Digital Olive 2 · 0⤊ 0⤋