Asked by Spencer
(cotxsec2x - cotx)/ (sinxtanx + cosx) = sinx
prove.
i keep getting (if i start from the left side) = cosx instead of sinx please help!
prove.
i keep getting (if i start from the left side) = cosx instead of sinx please help!
Answers
Answered by
Reiny
first of all you should have typed it correctly as
(cotxsec^2 x - cotx)/ (sinxtanx + cosx) = sinx
I read that first part as sec(2x), and wasted about 10 minutes trying to prove that.
so...
LS = [cotx(1/cos^2x - 1)]/[sin^2x/cosx + cosx]
= [cotx(1 - cos^2x)/cos^2x]/[(sin^x + cos^2x)cosx]
= [cotx(sin^2x/cos^2x)]/[1/cosx]
= [sinx/cosx]/[1/cosx]
= sinx
= RS
(cotxsec^2 x - cotx)/ (sinxtanx + cosx) = sinx
I read that first part as sec(2x), and wasted about 10 minutes trying to prove that.
so...
LS = [cotx(1/cos^2x - 1)]/[sin^2x/cosx + cosx]
= [cotx(1 - cos^2x)/cos^2x]/[(sin^x + cos^2x)cosx]
= [cotx(sin^2x/cos^2x)]/[1/cosx]
= [sinx/cosx]/[1/cosx]
= sinx
= RS
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