given that $sec\theta =-\frac{37}{12}$, what is the value of $cot\theta$, for $\frac{pi}{2}<\theta<pi$?\n$-\f…

Explanation:

Step1: Recall secant - cosine relationship

Since $\sec\theta=\frac{1}{\cos\theta}=-\frac{37}{12}$, then $\cos\theta =-\frac{12}{37}$.

Step2: Use Pythagorean identity

We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. Substitute $\cos\theta=-\frac{12}{37}$ into it: $\sin^{2}\theta=1 - \cos^{2}\theta=1-\left(-\frac{12}{37}\right)^{2}=1-\frac{144}{1369}=\frac{1369 - 144}{1369}=\frac{1225}{1369}$. Since $\frac{\pi}{2}<\theta<\pi$, $\sin\theta>0$, so $\sin\theta=\frac{35}{37}$.

Step3: Recall cotangent - sine/cosine relationship

$\cot\theta=\frac{\cos\theta}{\sin\theta}$. Substitute $\cos\theta =-\frac{12}{37}$ and $\sin\theta=\frac{35}{37}$ into it: $\cot\theta=\frac{-\frac{12}{37}}{\frac{35}{37}}=-\frac{12}{35}$.

Answer:

$-\frac{12}{35}$