Question

if $cos(\theta)=\frac{sqrt{42}}{7}$ and $0^{circ}<\theta<90^{circ}$, what is $\tan(\theta)$? write your answer in simplified, rationalized form. $\tan(\theta)=$

Explanation:

Step1: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

We know $\cos\theta=\frac{\sqrt{42}}{7}$, so $\sin^{2}\theta=1 - \cos^{2}\theta$. Substitute $\cos\theta$ into the formula: $\sin^{2}\theta=1-(\frac{\sqrt{42}}{7})^{2}=1-\frac{42}{49}=\frac{49 - 42}{49}=\frac{7}{49}=\frac{1}{7}$. Since $0^{\circ}<\theta<90^{\circ}$, $\sin\theta>0$, so $\sin\theta=\sqrt{\frac{1}{7}}=\frac{\sqrt{7}}{7}$.

Step2: Use the formula for $\tan\theta$

The formula for $\tan\theta$ is $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Substitute $\sin\theta=\frac{\sqrt{7}}{7}$ and $\cos\theta=\frac{\sqrt{42}}{7}$ into it: $\tan\theta=\frac{\frac{\sqrt{7}}{7}}{\frac{\sqrt{42}}{7}}=\frac{\sqrt{7}}{\sqrt{42}}$. Rationalize the denominator: $\frac{\sqrt{7}}{\sqrt{42}}\times\frac{\sqrt{42}}{\sqrt{42}}=\frac{\sqrt{7\times42}}{42}=\frac{\sqrt{7\times6\times7}}{42}=\frac{7\sqrt{6}}{42}=\frac{\sqrt{6}}{6}$.

Answer:

$\frac{\sqrt{6}}{6}$