Question

Prove the identity. tan theta /(cos theta - sec theta) = -csc theta

Answer 1

sin^2 theta + cos^2 theta = 1 was an identity I was using

Answer 2

OK. Start by putting it in terms of sine and cosine. That will let you do some workk with the fraction on the right.

Answer 3

my friend was able to solve the question but she let the denominator stay the same but changed all numerators into sin/cos.. is that ok?

Answer 4

\[\frac{\frac{sin}{cos}}{cos-sec}\] is where I started.

Answer 5

tan theta/(cos theta - sec theta)=(sin theta /cos theta)/(cos theta - 1/cos theta)=(sin theta/cos theta)/{(cos square theta -1)}/cos theta=sin theta/{-(1-cos square theta)}=-sin theta/sin square theta= - cosec theta

Answer 6

I was done in 7 more steps....

Answer 7

i got to the part where it's sin theta/cos theta = -cos^2theta +1 am i on the right track?

Answer 8

You have to work all on one side of the = sign for a proof of a trig function.

Answer 9

if i get rid of a denominator on one side.. wouldn't i have to multiply the other side by it as well?

Answer 10

You can't work proofs that way. You have to stay with things that don't need to move across the = sign. Like multiplication by 1. Well, 1 has a lot of faces. For example, all of these are 1:\[sin^2+cos^2\\ \frac{sin}{sin}\\ cot\cdot tan\]

Answer 11

yes u r on the right side

Answer 12

i am still so confused.. can you do the question from start to finish so i can see the work please

Answer 13

rahulon did.

Answer 14

Let me reformat some.

Answer 15

\[\frac{\tan \theta}{\cos \theta - \sec \theta} \Rightarrow \]The first step:
\[\frac{\frac{\sin \theta}{\cos theta}}{\cos \theta - \frac{1}{\cos \theta}} \Rightarrow \]

Answer 16

Now, you have a fraction divided by sometthing, that can be changed to multiplication and simplified. That is the big key here.

Answer 17

ok i understand that

Answer 18

does that give u sintheta cos theta / cos ^2 theta

Answer 19

This is where he and I did different things, but we both get to the same end. More of his:\[\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\cos^2 \theta -1}{\cos \theta}}\]

Answer 20

k

Answer 21

\[\frac{\sin \theta}{-(1-\cos^2 \theta)}\]

Answer 22

\[\frac{-\sin \theta}{\sin^2 \theta}\]

Answer 23

\[-\csc \theta \]That was his way, mine, without all the pretty prenting reformats was:
tan/(cos-sec)=-csc
(sin/cos)/(cos-sec)
(sin/cos)*1/(cos-sec)
sin/(cos(cos-sec))
sin/(cos^2-[cos(1/cos)])
sin/(cos^2-1)
sin/-sin^2
1/-sin
-csc

Answer 24

Same end result and we never worled the right side. We stayed on the left.

Answer 25

He also put together a step or two that I show, no big difference, no simpler or harder. He just went for cos earlier than I did.

Answer 26

thank you!!!!!! i understand now!