Question

assume that θ is an acute angle in a right triangle satisfying the given condition. evaluate the remaining trigonometric functions. cosθ = 5/13 cscθ = 13/12 (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) secθ = 13/5 (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) cotθ = (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall Pythagorean identity

In a right - triangle, if $\cos\theta=\frac{5}{13}$, let the adjacent side $a = 5$ and the hypotenuse $c = 13$. Using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, we find the opposite side $b$. So $b=\sqrt{c^{2}-a^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}=12$.

Step2: Find $\sin\theta$

By the definition of sine in a right - triangle $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, so $\sin\theta=\frac{12}{13}$.

Step3: Find $\csc\theta$

Since $\csc\theta=\frac{1}{\sin\theta}$, then $\csc\theta=\frac{13}{12}$.

Step4: Find $\sec\theta$

As $\sec\theta=\frac{1}{\cos\theta}$, and $\cos\theta=\frac{5}{13}$, so $\sec\theta=\frac{13}{5}$.

Step5: Find $\tan\theta$

Using the formula $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$.

Step6: Find $\cot\theta$

Since $\cot\theta=\frac{1}{\tan\theta}$, then $\cot\theta=\frac{5}{12}$.

Answer:

$\sin\theta=\frac{12}{13}$, $\csc\theta=\frac{13}{12}$, $\sec\theta=\frac{13}{5}$, $\tan\theta=\frac{12}{5}$, $\cot\theta=\frac{5}{12}$