Question

use trigonometric identities to simplify:
(csc(x)-tan(x))sin(x)cos(x)

Answer 1

write csc and tan in terms of sin and cos

Answer 2

csc =??
tan= ??
what is reciprocal of tan and csc ?

Answer 3

reciprocal of tanx is 1/cotx
reciprocal of cscx is 1/sinx
?

Answer 4

@Nnesha

Answer 5

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

Answer 6

can you do that ?? :)

Answer 7

@Nnesha So the common denom would be sin(x)cos(x) right?

Answer 8

yep

Answer 9

\[(\color{Red}{\frac{ ??-?? }{ \sin(x)cos(x)} })\sin(x)\cos(x)\]
multiply top of first fraction by the bottom of the 2nd fraction
multiply numerator of 2nd fraction by denominator of first fraction
\[(\color{Red}{\frac{ 1(cos(x))-sin(x)sin(x) }{ \sin(x)cos(x)} })\sin(x)\cos(x)\]

Answer 10

try to simplify tht from there

Answer 11

@Nnesha Would I cancel the denominator with the outside?

Answer 12

yes you can

Answer 13

@Nnesha so would my final answer be \[\cos(x)-\sin^{2}(x)\]

Answer 14

you can simplify that sin^2 = ?
btw do you have options ?

Answer 15

@Nnesha no I don't
\[\sin ^{2}x = \frac{ 1-\cos2(x) }{ 2 }\]

Answer 16

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

Answer 17

but i don't think we should substitute sin^2
just leave it as cos(x)-sin^2(x)

Answer 18

oh okay then thanks

Answer 19

my pleasure