Question

Verify the trig identity:
cot t + sin t/ 1 +cos t = csc t

Answer 1

\[\cot (t)+\sin(\frac{ t }{ 1 })+\cos(t)= \csc(t)\]

Answer 2

\[\csc (t)=\frac{ 1 }{ \sin(t) }\]
\[\cot (t)= \frac{ \cos(t) }{ \sin(t) }\]

Answer 3

Sorry, I still don't get it.

Answer 4

\[\frac{ \cos(t) }{ \sin (t) }+\sin(t) +\cos (t)= \frac{ 1 }{ \sin(t) }\]

Answer 5

common denominator is sin(t)

Answer 6

\[\frac{ \cos(t) }{ \sin(t) }+\frac{ \sin(t)*\sin(t) }{ \sin(t) }+\frac{ \cos(t)*\sin(t) }{ \sin(t) }=\]

Answer 7

something wring with the equation ?!

Answer 8

i couldn't come up with the answer, so i evaluate both right and left sides they are not not equal

Answer 9

\[\cot(t)+\frac{\sin(t)}{1+\cos(t)}=\csc(t)\]
I'm assuming that's what you meant? You should REALLY add parenthesis to make things clearer:
cot(t) + sin(t)/[1 +cos(t)] = csc(t)

Answer 10

\[\frac{\cos(t)}{\sin(t)}+\frac{\sin(t)}{1+\cos(t)}=\frac{1}{\sin(t)}\]

Answer 11

Multiply by sin(t)

Answer 12

\[\cos(t)+\frac{\sin^2(t)}{1+\cos(t)} = 1\]

Answer 13

multiply by 1+cos(t)

Answer 14

\[\cos(t)[1+\cos(t)]+\sin^2(t)=1[1+\cos(t)]\]

Answer 15

\[\cos(t)+\cos^2(t)+\sin^2(t)=1+\cos(t)\]

Answer 16

trig identity: \[\sin^2(t)+\cos^2(t)=1\]

Answer 17

Therefore:
\[\cos(t)+1 = 1+\cos(t)\]

Answer 18

Just a reminder, order of operations is very important - you can't just randomly throw away parenthesis an expect people to know how to read your equation

Answer 19

I thought that the problem was intended to be:

(cot t + sin t)/(1+cos t) = csc t which I don't think is an identity.

Answer 20

That could be the case; I changed the order of operations in the given problem to what I think it should've been, but as it's written, like you said, the equality does not hold.