Using Trigonometric Identities.
2007-07-29 07:47:43 · 6 answers · asked by Bobby 1 in Science & Mathematics ➔ Mathematics
LHS (brackets required)
(cos ² x - sin ² x) / (1 - sin ² x / cos ² x)
(cos ² x - sin ² x) / [cos ² x - sin ² x] / cos ² x
cos ² x (cos ² x - sin ² x) / (cos ² x - sin ² x)
cos ² x
RHS
cos ² x
LHS = RHS
2007-07-29 08:40:27 · answer #1 · answered by Como 7 · 0⤊ 0⤋
Cos^2-Sin^2 / ( 1-Tan^2 = Cos^2
Tan^2 =( Sin^2 / Cos^2) = Cos^2............. rewrite
Cos^2-Sin^2 / (1-(Sin^2/Cos^2)) = Cos^2.......Rewrite
Cos^2-Sin^2 / (( Cos^2-Sin^2) / Cos^2) + Cos^2 ... Rewrite
[Cos^2( Cos^2-Sin^2)]/( Cos^2-Sin^2) = Cos^2 ...Divide
Cos^2 = Cos^2 Equivalent
2007-07-29 08:26:17 · answer #2 · answered by ? 3 · 0⤊ 0⤋
sin-a million is the inverse functionality of sin In tiers, this means that considering the fact that sin(ninety) = a million, it follows that sin-a million(a million) = ninety you utilize those applications once you recognize from some computation what sin, cos or tan of an perspective is, yet you are able to desire to compute the attitude itself, no longer its sin, cos or tan cost.
2016-10-09 12:36:14 · answer #3 · answered by Erika 3 · 0⤊ 0⤋
cos^2=1-sin^2
tan^2=sin^2/cos^2
so...
[cos^2-sin^2] / [1 - sin^2/cos^2]
[cos^2-sin^2] / [cos^2-sin^2 / cos^2]
reduces to cos^2
2007-07-29 07:55:23 · answer #4 · answered by hrhbg 3 · 0⤊ 0⤋
yeah so??? tan= sin / cos
substitue in denominator n solve!
2007-07-29 07:56:35 · answer #5 · answered by harry_g12002 1 · 0⤊ 1⤋
cheese
2007-07-29 07:51:10 · answer #6 · answered by Anonymous · 0⤊ 1⤋