Answer
$$\sin\theta=\frac{2\sqrt6}{5}$$
$$\sec\theta=5$$
$$\csc\theta=\frac{5\sqrt6}{12}$$
$$\tan\theta=2\sqrt6$$
$$\cot\theta=\frac{\sqrt6}{12}$$
Work Step by Step
$$\cos\theta=\frac{1}{5}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant I}$$
1) Reciprocal Identities:
$$\sec\theta=\frac{1}{\cos\theta}$$
$$\sec\theta=\frac{1}{\frac{1}{5}}=5$$
2) Pythagorean Identities:
$$\tan^2\theta+1=\sec^2\theta$$
$$\tan^2\theta=\sec^2\theta-1$$
$$\tan^2\theta=5^2-1=25-1=24$$
$$\tan\theta=\pm\sqrt{24}=\pm2\sqrt6$$
3) Quotient Identities:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
We know that $\theta$ is in quadrant I, where both $\sin\theta\gt0$ and $\cos\theta\gt0$. Therefore, with the above identity, we can conclude that $\tan\theta\gt0$.
$$\tan\theta=2\sqrt6$$
$$\frac{\sin\theta}{\cos\theta}=2\sqrt6$$
$$\sin\theta=(2\sqrt6)\times\cos\theta=(2\sqrt6)\times\frac{1}{5}=\frac{2\sqrt6}{5}$$
4) Reciprocal Identities:
$$\tan\theta=\frac{1}{\cot\theta}$$
$$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{2\sqrt6}=\frac{\sqrt6}{2\times6}=\frac{\sqrt6}{12}$$
Also, $$\csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{2\sqrt 6}{5}}=\frac{5}{2\sqrt 6}=\frac{5\sqrt 6}{12}$$
