Trigonometry 7th Edition

$\sin\theta$ = $\frac{1}{2}$ $\cos\theta$ = - $\frac{\sqrt 3}{2}$ $\tan\theta$ = - $\frac{1}{\sqrt 3}$ $\sec\theta$ = - $\frac{2}{\sqrt 3}$ $\cot\theta$ = - $\sqrt 3$
Given $\csc\theta$ = 2 By reciprocal identity- $\sin\theta$ = $\frac{1}{\csc\theta}$ = $\frac{1}{2}$ We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ Given $\csc\theta$ is positive and $\cos\theta$ is negative, therefore $\theta$ terminates in Q II. $\cos\theta$ = - $\sqrt (1-\sin^{2}\theta)$ substitute the value of $\sin\theta$- $\cos\theta$ = - $\sqrt (1-(\frac{1}{2})^{2})$ $\cos\theta$ = - $\sqrt (1-\frac{1}{4})$ $\cos\theta$ = - $\sqrt (\frac{4-1}{4})$ = $\sqrt (\frac{3}{4})$ $\cos\theta$ = - $\frac{\sqrt 3}{2}$ By ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ Substituting the values of $\sin\theta$ and $\cos\theta$- $\tan\theta$ = $\frac{1/2}{ -\sqrt 3/2}$ = - $\frac{1}{\sqrt 3}$ From reciprocal identities- $\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{-\sqrt 3/2}$ = - $\frac{2}{\sqrt 3}$ $\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$