## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{\sqrt{17}}{17}$$ $$\cos\theta=\frac{4\sqrt{17}}{17}$$ $$\cot\theta=-4$$ $$\sec\theta=\frac{\sqrt{17}}{4}$$ $$\csc\theta=-\sqrt{17}$$
$$\tan\theta=-\frac{1}{4}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant IV}$$ 1) Reciprocal Identities: $$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{-\frac{1}{4}}=-4$$ 2) Pythagorean Identities: $$\sec^2\theta=\tan^2\theta+1=(-\frac{1}{4})^2+1=\frac{1}{16}+1=\frac{17}{16}$$ $$\sec\theta=\pm\frac{\sqrt{17}}{4}$$ Also, $$\csc^2\theta=\cot^2\theta+1=(-4)^2+1=17$$ $$\csc\theta=\pm\sqrt{17}$$ 3) Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}\hspace{1cm}\text{and}\hspace{1cm}\csc\theta=\frac{1}{\sin\theta}$$ We know that $\theta$ is in quadrant IV, where $\sin\theta\lt0$ but $\cos\theta\gt0$. Therefore, with the above identities, we can conclude that $\sec\theta\gt0$ but $\csc\theta\lt0$. That means, $$\sec\theta=\frac{\sqrt{17}}{4}\hspace{1cm}\text{and}\hspace{1cm}\csc\theta=-\sqrt{17}$$ We can calculate $\sin\theta$ and $\cos\theta$ here, too: $$\sin\theta=\frac{1}{\csc\theta}=\frac{1}{-\sqrt{17}}=-\frac{\sqrt{17}}{17}$$ $$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\frac{\sqrt{17}}{4}}=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}$$