given the value of one trigonometric function of an acute angle $\theta$, find the values of the remaining five trigonometric functions of $\theta$.
23) $\sin\theta=\frac{2}{\sqrt{13}}$
$\cos\theta =$
$\tan\theta =$
$\csc\theta =$
$\sec\theta =$
$\cot\theta =$

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Accepted Answer

# Explanation:
## Step1: Find $\cos	heta$ using $\sin^{2}	heta+\cos^{2}	heta = 1$
Given $\sin	heta=\frac{2}{\sqrt{13}}$, then $\sin^{2}	heta=\frac{4}{13}$.
Substitute into the identity: $\frac{4}{13}+\cos^{2}	heta = 1$.
So, $\cos^{2}	heta=1 - \frac{4}{13}=\frac{13 - 4}{13}=\frac{9}{13}$.
Since $	heta$ is acute, $\cos	heta=\frac{3}{\sqrt{13}}$.
## Step2: Find $	an	heta$ using $	an	heta=\frac{\sin	heta}{\cos	heta}$
$	an	heta=\frac{\frac{2}{\sqrt{13}}}{\frac{3}{\sqrt{13}}}=\frac{2}{3}$.
## Step3: Find $\csc	heta$ using $\csc	heta=\frac{1}{\sin	heta}$
$\csc	heta=\frac{\sqrt{13}}{2}$.
## Step4: Find $\sec	heta$ using $\sec	heta=\frac{1}{\cos	heta}$
$\sec	heta=\frac{\sqrt{13}}{3}$.
## Step5: Find $\cot	heta$ using $\cot	heta=\frac{1}{	an	heta}$
$\cot	heta=\frac{3}{2}$.

# Answer:
$\cos	heta=\frac{3}{\sqrt{13}}$, $	an	heta=\frac{2}{3}$, $\csc	heta=\frac{\sqrt{13}}{2}$, $\sec	heta=\frac{\sqrt{13}}{3}$, $\cot	heta=\frac{3}{2}$

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Question

Explanation:

Step1: Find $\cos\theta$ using $\sin^{2}\theta+\cos^{2}\theta = 1$

Step2: Find $\tan\theta$ using $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Step3: Find $\csc\theta$ using $\csc\theta=\frac{1}{\sin\theta}$

Step4: Find $\sec\theta$ using $\sec\theta=\frac{1}{\cos\theta}$

Step5: Find $\cot\theta$ using $\cot\theta=\frac{1}{\tan\theta}$

Answer: