## Trigonometry 7th Edition

$\cos\theta$ = $\frac{5}{13}$ $\tan\theta$ = $\frac{12}{5}$ $\csc\theta$ = $\frac{13}{12}$ $\sec\theta$ = $\frac{13}{5}$ $\cot\theta$ = $\frac{5}{12}$
We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ As $\theta$ terminates in Q I, Therefore $\cos\theta$ will be positive- $\cos\theta$ = $\sqrt (1-\sin^{2}\theta)$ substitute the given value of $\cos\theta$- $\cos\theta$ = $\sqrt (1-(\frac{12}{13})^{2})$ $\cos\theta$ = $\sqrt (1-\frac{144}{169})$ $\cos\theta$ = $\sqrt (\frac{169-144}{169})$ = $\sqrt (\frac{25}{169})$ $\cos\theta$ = $\frac{5}{13}$ By ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ Substituting the values of $\sin\theta$ and $\cos\theta$- $\tan\theta$ = $\frac{12/13}{5/13}$ = $\frac{12}{5}$ From reciprocal identities- $\csc\theta$ = $\frac{1}{\sin\theta}$ = $\frac{1}{12/13}$ = $\frac{13}{12}$ $\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{5/13}$ = $\frac{13}{5}$ $\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{12/5}$ = $\frac{5}{12}$ As $\theta$ terminates in Q I, Therefore all trigonometric functions will be positive.