Question

is this correct ? cot^2x +1/cot^2x=csc^2x/cos^2x/sin^2x=1/sin^2x/cos^2x/sin^2x=cos^2x

Answer 1

original question is cot^2x+1/cot^2x

Answer 2

Using your trig identities:
\[\cot^2 x+1=\csc^2 x\]
\[\frac{\cot^2 x+1}{\cot^2 x}=\frac{\csc^2 x}{\cot^2 x}\]
\[=\frac{1}{\sin^2 x} \times \frac{\sin^2 x }{\cos^2 x}\]
\[=\frac{1}{\cos^2 x}\]
\[=\csc^2 x\]

Answer 3

so it reverst back to csc^2x?

Answer 4

NO...
\[\huge \frac{\csc^2}{\cot^2 x}=\frac{\frac{1}{\sin^2 x}}{\frac{1}{\tan^2 x}}\]
Correct?

Answer 5

but doesnt cotx = cosx/sinx?

Answer 6

thats what Ihave in my notes..

Answer 7

Separate into two fractions:
\[\huge \frac{\frac{1}{\sin^2 x}}{\frac{1}{\tan^2 x}}=\frac{1}{\sin^2 x}\div \frac{1}{\tan^2 x}\]
\[\huge =\frac{1}{\sin^2 x} \div \frac{\cos^2 x}{\sin^2 x}\]
When you multiply you switch one fraction and get it's reciprocal.
\[\huge =\frac{1}{\sin^2 x}\times \frac{\sin^2 x}{\cos^2 x}\]

Answer 8

You didn't read your notes on the basics of changing a division sign into a multiplication sign it seems.

Answer 9

which equal cos^2x

Answer 10

How does:
\[\frac{1}{\cos^2 x}\]
equal
\[\cos^2 x\]

Answer 11

it doesn't thats just what my brain saw and computed

Answer 12

So do you get it now that cos^2 x is not the answer?

Answer 13

yeah I see where I went wrong.

Answer 14

thank you

Answer 15

No worries.