Question

cotx(cosx -1) wrting this in terms of sinx

cosx/sinx * cosx-1/1/secx is what i got so far but not sure if I am in ball park or if I am off with the secx?

Answer 1

im thinking my answer would be secx and that the cosx would cancel out but not sure

Answer 2

Your question says to write it in terms of sin x. sec x is not in terms of sin x.....

Answer 3

\[\cot x(\cos x-1)=\frac{\cos x}{\sin x}(\cos x-1)\]
\[\large =\frac{\cos^2 x}{\sin x}-\frac{\cos x}{\sin x}\]
\[\large =\frac{\cos^2 x-\cos x}{\sin x}\]
Using the identity:
\[\cos^2 x+\sin^2 x=1\]
\[\cos^2 x=1-\sin^2 x\]
\[\cos x =\pm \sqrt{1-\sin^2 x}\]
\[\large =\frac{1-\sin^2 x-\cos x}{\sin x}\]
\[\large =\frac{1-\sin^2 x\pm \sqrt{1-\sin^2 x}}{\sin x}\]

Answer 4

so we factored the expressions in order to solve?

Answer 5

Where does your question say solve unless this is one part of a question.

Answer 6

it doesnt

Answer 7

So Voila! We're done here if you understand completely what I did there. If not, then you can ask away at anything that's mind-boggling about the question.

Answer 8

cos2xsinx−cosxsinx
I gues is what i don't understand is how you got this

Answer 9

I distributed/expanded to get rid of the brackets.

Answer 10

paranthesis*

Answer 11

oh ok I see it now
\thank you for your assistance

Answer 12

No worries.