Question

Trig/Precal 2 Please help

Answer 1

\[\frac{ 1+cosu }{ sinu } + \frac{ sinu }{ cosu+1 }\]

Answer 2

So far I have this:

\[\frac{ (1+sinu)^2 + \sin^2u }{ (cosu+1)(sinu) }\]

\[\frac{ \sin^2u+2sinu+1+\sin^2u }{ (cosu+1)(sinu) }\]

Answer 3

I'm supposed to Simplify the trigonometric expression.by the way.

Answer 4

You are correct in believing that the least common denominator of these two fractions is (sin u)(cos u + 1). But I see problems with your numerators.

Would you mind multiplying out (1 + cos u)(1 + cos u)?

Please multiply: (sin u)(sin u).

Write your results here, please.

Answer 5

\[
\frac{\cos (u)+1}{\sin (u)}+\frac{\sin
(u)}{\cos (u)+1}=\\\frac{\sin
^2(u)+(\cos (u)+1)^2}{\sin (u) (\cos
(u)+1)}=\\\frac{\sin ^2(u)+\cos ^2(u)+2
\cos (u)+1}{\sin (u) (\cos
(u)+1)}=
\]
Can you finish it now?

Answer 6

What is
\[
\sin^2(u) + \cos^2(u)
\]

Answer 7

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

Answer 8

OOH I messed up on the top part, its supposed to be (1+cos^u)^2
Thanks eliassa. You too mathmale. Thanks guys

Answer 9

YW

Answer 10

Your showing your work helped me help you. Glad you're making progress!

Answer 11

Wish I could have gave you both medals haha.