Question

find the exact value of each of the remaining trigonometric functions of \\( \theta \\).
\\( \sec \theta = 6, \quad \tan \theta > 0 \\)
\\( \sin \theta = \frac { \sqrt { 35 } } { 6 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
\\( \cos \theta = \frac { 1 } { 6 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
\\( \tan \theta = \sqrt { 35 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
\\( \cot \theta = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find $\cos\theta$

Use the identity $\sin^{2}\theta+\cos^{2}\theta = 1$.
Given $\sin\theta=\frac{\sqrt{35}}{6}$, then $\cos^{2}\theta=1 - \sin^{2}\theta$.
Substitute $\sin\theta$: $\cos^{2}\theta=1-\frac{35}{36}=\frac{36 - 35}{36}=\frac{1}{36}$.
Since $\sec\theta = 6>0$ and $\tan\theta>0$, $\cos\theta=\frac{1}{6}$ (because $\cos\theta=\frac{1}{\sec\theta}$ and in the first - quadrant where $\tan\theta>0$ and $\sec\theta>0$, cosine is positive).

Step2: Find $\tan\theta$

Use the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$.
Substitute $\sin\theta=\frac{\sqrt{35}}{6}$ and $\cos\theta=\frac{1}{6}$, then $\tan\theta=\frac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}=\sqrt{35}$.

Step3: Find $\cot\theta$

Use the identity $\cot\theta=\frac{1}{\tan\theta}$.
Since $\tan\theta=\sqrt{35}$, then $\cot\theta=\frac{1}{\sqrt{35}}=\frac{\sqrt{35}}{35}$ (rationalize the denominator: $\frac{1\times\sqrt{35}}{\sqrt{35}\times\sqrt{35}}$).

Answer:

$\cos\theta=\frac{1}{6}$, $\tan\theta=\sqrt{35}$, $\cot\theta=\frac{\sqrt{35}}{35}$