Question

Simplify the expression csc x- cosx cot x

Answer 1

HI!!!

Answer 2

easier to do the algebra using \(a\) for cosine and \(b\)n for sine, making this \[\frac{1}{b}-a\times \frac{a}{b}\]

Answer 3

the denominators are the same, so you subtract in the numerator and get \[\frac{1-a^2}{b}\]

Answer 4

now back to trig \[\frac{1-\cos^2(x)}{\sin^x)}=\frac{\sin^2(x)}{\sin(x)}\] cancel and you are done

Answer 5

I Don't understand hoe to cancel that

Answer 6

how

Answer 7

\[\frac{x^2}{x}=?\]

Answer 8

\[\frac{5^2}{5}=?\]

Answer 9

You're left with 5 right?

Answer 10

yes right

Answer 11

and \(\frac{x^2}{x}=?\)

Answer 12

X

Answer 13

right

Answer 14

so \[\frac{\sin^2(x)}{\sin(x)}=?\]

Answer 15

Sin

Answer 16

\[\color\magenta\heartsuit\]

Answer 17

I don't get why cos turned into sin

Answer 18

cosine did not turn in to sine

Answer 19

\[\sin^2(x)+\cos^2(x)=1\] always, so \[1-\cos^2(x)=\sin^2(x)\]

Answer 20

i.e. \[1-\cos^2(x)\] turned in to \(\sin^2(x)\)

Answer 21

Okay but why r we dividing both sides of the equation by sin

Answer 22

lets go slow here

Answer 23

\[\csc(x)=\frac{1}{\sin(x)}\] did you know that one?q

Answer 24

No is that a trigidentity

Answer 25

that is the definition of cosecant, the reciprocal of sine

Answer 26

here is the deal
without knowing the basic trig identities none of these problems are going to be possible to do
they will just look like gibberish
make sure you know what tangent, cotangent, secant, and cosecant are in terms of sine and cosine
without that, none are possible

Answer 27

then you need to know a couple other ones
like \[\sin^2(x)+\cos^2(x)=1\]
after that, it is all algebra

Answer 28

I do know them, just not used to them. :( but I am gaining some clarity, yes my teacher gave us some phrases to remember those like that one