DanielaJohana
  • DanielaJohana
use trigonometric identities to simplify: (csc(x)-tan(x))sin(x)cos(x)
Mathematics
chestercat
  • chestercat
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Nnesha
  • Nnesha
write csc and tan in terms of sin and cos
Nnesha
  • Nnesha
csc =?? tan= ?? what is reciprocal of tan and csc ?
DanielaJohana
  • DanielaJohana
reciprocal of tanx is 1/cotx reciprocal of cscx is 1/sinx ?

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DanielaJohana
  • DanielaJohana
Nnesha
  • Nnesha
yes right and also tan = sin over cos \[\huge\rm [\color{red}{\csc(x)-\tan(x)}]\sin(x)\cos(x)\] \[(\color{Red}{\frac{ 1 }{ \sin(x)} -\frac{ \sin(x) }{ \cos(x) }})\sin(x)\cos(x)\] now solve the parentheses find common denomiantor
Nnesha
  • Nnesha
can you do that ?? :)
DanielaJohana
  • DanielaJohana
@Nnesha So the common denom would be sin(x)cos(x) right?
Nnesha
  • Nnesha
yep
Nnesha
  • Nnesha
\[(\color{Red}{\frac{ ??-?? }{ \sin(x)cos(x)} })\sin(x)\cos(x)\] multiply top of first fraction by the bottom of the 2nd fraction multiply numerator of 2nd fraction by denominator of first fraction \[(\color{Red}{\frac{ 1(cos(x))-sin(x)sin(x) }{ \sin(x)cos(x)} })\sin(x)\cos(x)\]
Nnesha
  • Nnesha
try to simplify tht from there
DanielaJohana
  • DanielaJohana
@Nnesha Would I cancel the denominator with the outside?
Nnesha
  • Nnesha
yes you can
DanielaJohana
  • DanielaJohana
@Nnesha so would my final answer be \[\cos(x)-\sin^{2}(x)\]
Nnesha
  • Nnesha
you can simplify that sin^2 = ? btw do you have options ?
DanielaJohana
  • DanielaJohana
@Nnesha no I don't \[\sin ^{2}x = \frac{ 1-\cos2(x) }{ 2 }\]
Nnesha
  • Nnesha
sin^2x is just equal to 1-cos^2\[\rm cos^2 \theta +\sin^2 \theta =1\] so when you solve for sin^2theat you will get 1-cos^2
Nnesha
  • Nnesha
but i don't think we should substitute sin^2 just leave it as cos(x)-sin^2(x)
DanielaJohana
  • DanielaJohana
oh okay then thanks
Nnesha
  • Nnesha
my pleasure

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