I know that $\cot\theta = 4/3$ how do I find $\csc\theta$?
I tried to do $\csc^2\theta - \cot^2\theta = 1$
This gives me $\csc^2\theta = 1 + \cot^2\theta$
this gives $csc^2\theta = 9/9 + 16/9 = 27/9 = \sqrt{3}$
is this wrong? My book gives the answer as $5/3$
I can never go more than $2$ homework problems without getting stuck.
$\endgroup$ 13 Answers
$\begingroup$I don't see what is your problem here... You know that $csc^2 x= 1+\cot^2 x=1+\frac{16}{9}=\frac{25}{9}$. From here, you get $\csc x=\frac{5}{3}$. It's pure algebra. Just look at what you have and where do you want to get, and as I said, do not rush with computations, since I notice you make very many elementary mistakes.
$\endgroup$ 3 $\begingroup$If $\cot{(x)} = 4/3$, then we have this picture
<
/| / | 3 5 / | / | x----- 4Now compute.
$\endgroup$ 1 $\begingroup$Do you know "Soh-Cah-Toa"? That is sine is the opposite side over hypoteneuse, cosine adjacent over hypoteneuse, and tangent is opposite over adjacent. Then cotangent will be adjacent over hypoteneuse. Therefore you have a right triangle where the side adjacent to your angle is 4 and the side across from your angle is 3. Pythagorean Theorem says the hypoteneuse is 5. Therefore cosecant, which is hypoteneuse over opposite is 5/3.
$\endgroup$ 2
