can you please show me the process...
assuming all angles are positive acute angles transform the left side of the equation into the right side
sec β - cos β / cos β = tan^2 β
(sec β minus cos β over cos β equals tan squared β)
2007-12-08 15:49:16 · 5 answers · asked by pirlo 1 in Science & Mathematics ➔ Mathematics
THANK YOU!
very helpful =D
2007-12-08 16:12:07 · update #1
THANK YOU ALL!
very helpful =D
2007-12-08 16:12:17 · update #2
Trig, 20 years later, still makes me cry.
2007-12-08 15:58:26 · answer #1 · answered by keshequa87 6 · 0⤊ 0⤋
A quicker approach:
(sec β - cos β) / cos β
= sec^2 β - 1
= tan^2 β
2007-12-08 23:57:40 · answer #2 · answered by RobertJ 4 · 1⤊ 0⤋
LHS
= (sec β - cos β) / cos β
= (1/cos β - cos β) / cos β
= (1 - cos^2 β) / cos^2 β
= sin^2 β / cos^2 β
= tan^2 β
= RHS.
2007-12-08 23:58:34 · answer #3 · answered by Madhukar 7 · 1⤊ 0⤋
(sec β - cos β) / cos β = tan^2 β
= (1/cosβ - cosβ)/cosβ/cosβ
=(1-cos^2β)/cos^2β
=sin^2β/cos^2β
=tan^2β
2007-12-08 23:56:06 · answer #4 · answered by norman 7 · 1⤊ 0⤋
(1/cos(Beta) - cos(Beta))/cos(Beta) = tan^2(Beta)
(1/cos(Beta) - cos(Beta))/cos(Beta) = sin^2(Beta)/cos^2(Beta)
1/cos(Beta) - cos(Beta) = sin^2(Beta)/cos(Beta)
1 - cos^2(Beta) = sin^2(Beta)
1 = sin^2(Beta) + cos^2(Beta) ~> Correct
2007-12-08 23:53:57 · answer #5 · answered by Anonymous · 1⤊ 0⤋