Question

cot(theta)-tan(theta) over sin(theta) + cos(theta) equals csc(theta) - sec(theta)
how do I proof this?

Answer 1

Let's use x instead of theta

Answer 2

okay

Answer 3

\[\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x\]

Answer 4

Do I have the problem stated correctly?

Answer 5

yes :)

Answer 6

would I have to replace cot(x) with cos(x)/sin(x)

Answer 7

and tan(x) with sin(x)/cos(x)

Answer 8

\[(\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\sin x+\cos x})\times \frac{\sin x \cos x}{\sin x \cos x}=\frac{\cos ^2x-\sin ^2x}{\sin x \cos x(\sin x+\cos x)}\]

Answer 9

Can you get it from there?

Answer 10

sadly no?
can you explain how I can simplify it more please?

Answer 11

Factor the numerator and then the cosx + sinx will cancel.

Answer 12

I'm still lost :(

Answer 13

@alejandrop95 This is not your question

Answer 14

how would I factor it though? I'm sorry I am new to trig.

Answer 15

Can you factor this:
\[a^2-b^2\]

Answer 16

I don't think you can because it doesn't have the same coefficient

Answer 17

\[a^2−b^2=(a-b)(a+b)\]

Answer 18

\[\cos ^2x-\sin ^2x=(\cos x-\sin x)(\cos x+\sin x)\]

Answer 19

oh okay that makes sense. so would it be cos-sin(cos+sin)

Answer 20

yes

Answer 21

then that would cancel with the denominator correct?

Answer 22

Yes. sinx + cosx will cancel

Answer 23

which will leave 1/sin(x)+cos(x)

Answer 24

no

Answer 25

using the reciprocal identities this will equal csc(x)-sec(x) ?

Answer 26

cos(x)-sin(x)/sin(x)+cos(x0

Answer 27

this is where I would use the reciprocal identities?

Answer 28

the cos(x) cancel and the sin(x) cancels?