noobinmath

Hello,

I have the following expression:

1/((sinx)^2) - 4/3, and I want to get to the following expression:

1/((tgx)^2) - 1/3. I wasn't able to find the right identities, maybe you could help?

Thanks!

skeeter

$\dfrac{1}{\sin^2{x}} - 1 - \dfrac{1}{3}$

$(\csc^2{x} - 1) - \dfrac{1}{3}$

$\cot^2{x} - \dfrac{1}{3}$

$\dfrac{1}{\tan^2{x}} - \dfrac{1}{3}$

skipjack

1/sin²x - 4/3 = (cos²x + sin²x)/sin²x - 4/3 = cos²x/sin²x + 1 - 4/3 = 1/tan²x - 1/3

noobinmath

Thanks guys!!! Even got 2 options The more the merrier