Question

help?!

Answer 1

@alivejeremy hey again!

Answer 2

I don't get this one sorry! :l

Answer 3

But the question is true

Answer 4

@mathstudent55 @Michele_Laino

Answer 5

Thank you for trying!
Do you guys know? @Mehek14 @Michele_Laino @mathmale

Answer 6

what is the question

Answer 7

Its in the link!

Answer 8

I'm really sorry I don't know this one:(

Answer 9

Alrighty thanks anyways! :)

Answer 10

if we apply the definition of \(\cot \theta\) and \(\csc \theta\), we can rewrite the identity, as follows:
\[1 + \frac{{{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}} = \frac{1}{{{{\left( {\sin \theta } \right)}^2}}}\]

Answer 11

next, If I multiply both sides by \((\sin \theta)^2\), I get:
\[{\left( {\sin \theta } \right)^2}\left\{ {1 + \frac{{{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}}} \right\} = {\left( {\sin \theta } \right)^2}\frac{1}{{{{\left( {\sin \theta } \right)}^2}}}\]
please simplify

Answer 12

Did you mess up on the last part?

Answer 13

I got Sin^2(theta)+Cot^2(theta) ...o.O

Answer 14

i'm so done with this post c:

Answer 15

lmbooo @alivejeremy Me too!

Answer 16

lmao

Answer 17

after the multiplication, I get this:
\[{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1\]
@Allieeslabae