Question

let $\theta$ be an angle in quadrant i such that $cos\theta=\frac{7}{10}$. find the exact values of $csc\theta$ and $cot\theta$. $csc\theta =$ $cot\theta =$

Explanation:

Step1: Find $\sin\theta$ using identity

We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. Given $\cos\theta=\frac{7}{10}$, then $\sin^{2}\theta=1 - \cos^{2}\theta$. Substituting $\cos\theta$: $\sin^{2}\theta=1-(\frac{7}{10})^{2}=1-\frac{49}{100}=\frac{100 - 49}{100}=\frac{51}{100}$. Since $\theta$ is in quadrant I, $\sin\theta=\frac{\sqrt{51}}{10}$.

Step2: Calculate $\csc\theta$

Recall that $\csc\theta=\frac{1}{\sin\theta}$. Substituting $\sin\theta = \frac{\sqrt{51}}{10}$, we get $\csc\theta=\frac{10}{\sqrt{51}}=\frac{10\sqrt{51}}{51}$.

Step3: Calculate $\cot\theta$

Recall that $\cot\theta=\frac{\cos\theta}{\sin\theta}$. Substituting $\cos\theta=\frac{7}{10}$ and $\sin\theta=\frac{\sqrt{51}}{10}$, we have $\cot\theta=\frac{\frac{7}{10}}{\frac{\sqrt{51}}{10}}=\frac{7}{\sqrt{51}}=\frac{7\sqrt{51}}{51}$.

Answer:

$\csc\theta=\frac{10\sqrt{51}}{51}$
$\cot\theta=\frac{7\sqrt{51}}{51}$