Asked by Mel
How do I prove the following Identity?
sec x(sec x-cos x)=tan^2x
If there is a certain website or suggestion to help with these type of equations I would greatly appreciate it. I've been studying these for a while but still get pretty confused.
sec x(sec x-cos x)=tan^2x
If there is a certain website or suggestion to help with these type of equations I would greatly appreciate it. I've been studying these for a while but still get pretty confused.
Answers
Answered by
Anonymous
sec(x)*[sec(x)-cos(x)]=
sec(x)*sec(x)-sec(x)*cos(x)
Remark:
sec(x)=1/cos(x)
sec(x)*sec(x)-sec(x)*cos(x)=
1/cos(x) * 1/cos(x) - 1/cos(x) * cos(x)=
1/cos^2(x)-1=
1/cos^2(x) - cos^2(x)/cos^2(x)=
1-cos^2(x)/cos^2(x)
Remark:
sin^2(x)+cos^2(x)=1
sin^2(x)=1-cos^2(x)
1-cos^2(x)/cos^2(x) = sin^2(x)/cos^2(x) = tan^2(x)
sec(x)*sec(x)-sec(x)*cos(x)
Remark:
sec(x)=1/cos(x)
sec(x)*sec(x)-sec(x)*cos(x)=
1/cos(x) * 1/cos(x) - 1/cos(x) * cos(x)=
1/cos^2(x)-1=
1/cos^2(x) - cos^2(x)/cos^2(x)=
1-cos^2(x)/cos^2(x)
Remark:
sin^2(x)+cos^2(x)=1
sin^2(x)=1-cos^2(x)
1-cos^2(x)/cos^2(x) = sin^2(x)/cos^2(x) = tan^2(x)
Answered by
Reiny
LS = sec x(sec x-cos x)
= (1/cosx)(1/cosx - cosx)
= 1/cos^2x - 1
= sec^2 - 1
= tan^2x (by definition)
= RS
= (1/cosx)(1/cosx - cosx)
= 1/cos^2x - 1
= sec^2 - 1
= tan^2x (by definition)
= RS
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