Question

Simplify.
csc^2x-1 / cot^2x

Answer 1

\[\frac{ \csc^2x-1 }{ \cot^2x }\]

Answer 2

@mathstudent55

Answer 3

csc^2(x-1)tan^2(x)

Answer 4

we have to use these identities:
\[\large \begin{gathered}
\csc x = \frac{1}{{\sin x}} \hfill \\
\cot x = \frac{{\cos x}}{{\sin x}} \hfill \\
\end{gathered} \]

Answer 5

@jaysabelle hope this helps :3

Answer 6

Wait, I don't get it . . . Ugh, I hate trigonometry :/

Answer 7

\[\csc^2(x-1)\tan^2(x)\]

Answer 8

this is that correct term yeah its hard but mathway.com can help a lot ^_^

Answer 9

hint:
\[\large \frac{{{{\left( {\csc x} \right)}^2}}}{{{{\left( {\cot x} \right)}^2}}} = \frac{{\frac{1}{{{{\left( {\sin x} \right)}^2}}}}}{{\frac{{{{\left( {\cos x} \right)}^2}}}{{{{\left( {\sin x} \right)}^2}}}}} = \frac{1}{{{{\left( {\sin x} \right)}^2}}} \times \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 10

@timbo5lice So, you're saying that \[\csc^2(x-1)\tan^2(x)\] is the most simplified it cab be?

Answer 11

yes ^_^

Answer 12

oops..
\[\large \frac{{{{\left( {\csc x} \right)}^2} - 1}}{{{{\left( {\cot x} \right)}^2}}} = \frac{{\frac{1}{{{{\left( {\sin x} \right)}^2}}} - 1}}{{\frac{{{{\left( {\cos x} \right)}^2}}}{{{{\left( {\sin x} \right)}^2}}}}} = \frac{{1 - {{\left( {\sin x} \right)}^2}}}{{{{\left( {\sin x} \right)}^2}}} \times \frac{{{{\left( {\sin x} \right)}^2}}}{{{{\left( {\cos x} \right)}^2}}} = ...?\]

Answer 13

@Michele_Laino could you help me understand more of it? Trig really is my weakest subject.

Answer 14

can you continue from my last step?

Answer 15

No, I don't know how to go about it

Answer 16

try mathway.com its a calculator for this stuff (:

Answer 17

hint: