Question

Simplifying Trig Expressions Help?

Answer 1

\[\frac{ \cos ^{2}\theta }{\sin ^{2}\theta } + \csc \theta \sin \theta\]

Answer 2

I'm so confused, I got cot^2 but that is wrong

Answer 3

\( \dfrac{ \cos ^{2}\theta }{\sin ^{2}\theta } + \csc \theta \sin \theta\)

\( = \cot ^{2}\theta + \dfrac{1}{\sin \theta}\times \sin \theta \)

\( = \cot ^{2} \theta + 1 \)

Answer 4

ooohh the pythagoren identity~ ! :D

Answer 5

so it = csc^2 theta?

Answer 6

Now look at this:

\(sin^2 \theta + \cos ^2 \theta = 1\)

\( \dfrac{\sin^2 \theta}{sin^2 \theta} + \dfrac{\cos ^2 \theta}{\sin^2 \theta} = \dfrac{1}{\sin^2 \theta} \)

\(1 + \cot^2 \theta = \csc^2 \theta \)

Therefore, you can replace \(\cot^2 \theta\ + 1 \) with \(\csc^2 \theta\)

Answer 7

Yes, you got it!

Answer 8

Thanks so much :)

Answer 9

You're welcome.