Question

1+cot^2(-theta) = csctheta
I'm not allowed to change the right hand side of the equation

How would I go about doing this problem?

Answer 1

\[1+\cot ^2(-\theta)=\csc^2\theta \]

Answer 2

we can use this identity:
\[\Large \cot \left( { - \theta } \right) = \frac{{\cos \left( { - \theta } \right)}}{{\sin \left( { - \theta } \right)}} = \frac{{\cos \theta }}{{ - \sin \theta }} = - \frac{{\cos \theta }}{{\sin \theta }}\]

Answer 3

@minisweet4

Answer 4

How would that get me to \[\csc^2\theta\] in the end?

Answer 5

You have to do some arithmetic to get there.

Answer 6

hint:
if we apply that identity, we get:
\[\large1 + {\left[ {\cot \left( { - \theta } \right)} \right]^2} = 1 + {\left( { - \frac{{\cos \theta }}{{\sin \theta }}} \right)^2} = 1 + \frac{{{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}}\]

Answer 7

please simplify this expression:
\[\large 1 + {\left[ {\cot \left( { - \theta } \right)} \right]^2} = 1 + {\left( { - \frac{{\cos \theta }}{{\sin \theta }}} \right)^2} = 1 + \frac{{{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}} = ...?\]

Answer 8

furthermore, we can rewrite the right side of your expression, as below:
\[\Large {\left( {\csc \theta } \right)^2} = \frac{1}{{{{\left( {\sin \theta } \right)}^2}}}\]

Answer 9

I see what you did now

Answer 10

That's very good!

Answer 11

how'd you get rid of the negative in front of cos/sin?

Answer 12

The square got rid of it.

Answer 13

since, we can write this:
\[\large {\left( { - \frac{{\cos \theta }}{{\sin \theta }}} \right)^2} = \left( { - \frac{{\cos \theta }}{{\sin \theta }}} \right) \times \left( { - \frac{{\cos \theta }}{{\sin \theta }}} \right) = \frac{{{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\sin \theta } \right)}^2}}}\]

Answer 14

Oh okay, thank you both very much. I may be putting up another problem after this one

Answer 15

ok!

Answer 16

Thank you!