Question

Since I can't type theta I wrote it as x.

1+sinx+cosx/1+sinx-cosx=1+cosx/sinx

Prove that they're equal.

Answer 1

I started with the left side. I multiplied both top and bottom by (1+sinx)+cosx.

\[((1+\sin \theta)+\cos \theta)/((1+\sin \Theta)-\cos \theta) * (1+\sin \theta)+\cos \theta/(1+\sin \theta)+\cos \theta\]

Answer 2

Well wait, is it

\[\large \frac{1 + sin(x) + cos(x)}{1 + sin(x) - cos(x)} = \frac{1 + cos(x)}{sin(x)}\]

Answer 3

Yes that is correct.

Answer 4

I would just cross multiply that

\[\large sin(x)(1 + sin(x) + cos(x)) = (1 + cos(x))(1 + sin(x) - cos(x))\]

So then we would have

\[\large sin(x) + sin^2(x) + cos(x)sin(x) = 1 - cos^2(x) + sin(x) + cos(x)sin(x)\]

Answer 5

Got cut off a bit there, that last bit is \(\large sin(x)\)

Okay so now...lets simplify the right hand side a bit

We know \(\large cos^2(x) = 1 - sin^2(x)\) so

\[\large sin(x) + sin^2(x) + cos(x)sin(x) = 1 - (1 - sin^2(x)) + sin(x) + cos(x)sin(x)\]

Answer 6

I guess I'll just write the right hand side since that's all we need to work on

\[\large 1 - (1 - sin^2(x)) + sin(x) + cos(x)sin(x)\]
\[\large 1 + sin^2(x) - 1 + sin(x) + cos(x)sin(x)\]
\[\large sin^2(x) + sin(x) + cos(x)sin(x)\]

They match!

Answer 7

Wow thank you very much johnweldon1993! This method feels more efficient than the one I was trying. After all this is a proportion. Thanks again for the help! This will help me with my test coming up in later today.

Answer 8

Yeah I was doing the way you had above, and it was seeming monotonous, regardless no problem!

Answer 9

Next time I will try out this method instead. Goodbye for now and have a great day!

Answer 10

You as well!