Question

Prove the Trigonometric Equation
sinx/1-cosx +sinx/1+cosx =2cscx

Answer 1

I currently have sinx + sinx/1+cosx

Answer 2

\[\Large \frac{ \sin \theta }{ 1-\cos \theta }+\frac{\sin \theta}{1+\cos \theta}=2\csc \theta\] correct?

Answer 3

Yes

Answer 4

Alright, does it matter if you make the RHS look like the LHS or do you have to simplify the LHS to the RHS

Answer 5

I have to simplify the RHS to the LHS because it's more complex

Answer 6

I mean LHS to the RHS

Answer 7

Alright, well a tip is to work on both sides, get one side simplified to a point and then have the other side "catch up" to it. Anyways this looks simple enough, we just have to go and add the fractions so by getting like denominators we get
\[\Large \frac{\sin \theta(1+\cos \theta)+\sin \theta(1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}\]

Answer 8

I'm lost

Answer 9

Alright, well you know to add \[\Large \frac{a}{b}+\frac{c}{d}\] that you have to get both fractions with like denominators correct?

Answer 10

Yes

Answer 11

Then you can add the numerators

Answer 12

Never mind I got you

Answer 13

OK, well committ this to memory cause it makes these^^^ a lot faster \[\Large \frac{a}{b}+\frac{c}{d}=\frac{a \cdot d}{b \cdot d}+\frac{c \cdot b}{d \cdot b}=\frac{ad+cb}{db}\]

Answer 14

Okay sorry, it's just looking at that makes me confused carry on

Answer 15

Yeah just take the numerator of the first and multiply it by the denomin of the second then do that for the other fraction and then multiply the denominators, it saves time in simplifying them and then adding them.

Anyways we get \[ \Large \frac{\sin \theta+\sin \theta \cos \theta+\sin \theta-\sin \theta\cos \theta}{1-\cos^2 \theta}\]

Answer 16

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

Answer 17

Yeah, from there I got 2sin^2x/1-cos^2x

Answer 18

2sin theta/sin ^2 theta

Answer 19

I think you can simplify from there

Answer 20

2*1/sin theta

Answer 21

I'm sorry I don't know what to do next

Answer 22

1/sin theta=csc theta

Answer 23

The sin in the numerator cancels out with one of the sin in the denom.

Answer 24

1-cos^2 theta=sin^2 theta use this identity in the denominator

Answer 25

\[\Large \frac{2\sin \theta}{\sin^2 \theta}= \frac{2\sin \theta}{\sin \theta \cdot \sin \theta}=\frac{2}{\sin \theta}=2 \cdot \frac{1}{\sin \theta}= 2\csc \theta\]

Answer 26

Thanks a bunch dude

Answer 27

Yw