Answer
$\cos\theta$ = $\frac{5}{13}$
$\tan\theta$ = $\frac{12}{5}$
$\csc\theta$ = $\frac{13}{12}$
$\sec\theta$ = $\frac{13}{5}$
$\cot\theta$ = $\frac{5}{12}$
Work Step by Step
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.
