Answer
$\sin\theta$= $- \frac{\sqrt {95}}{12}$
$\cos\theta$ =$ \frac{7}{12}$ (given)
$\tan\theta$ =$- \frac{\sqrt {95}}{7}$
$\cot\theta$ =$- \frac{7}{\sqrt {95}}$
$\sec\theta$ =$ \frac{12}{7}$
$\csc\theta$ =$- \frac{12}{\sqrt {95}}$
Work Step by Step
Considering $\theta$ as reference angle, we may draw a right triangle ABC to calculate other trigonometric functions-
In right triangle ABC-
Given $\cos\theta$ =$ \frac{x}{r}$ = $ \frac{7}{12}$ (Ignoring sign)
Using Pythagorous theorem-
$y =\sqrt {r^{2}-x^{2}}$ =$ \sqrt {12^{2}-7^{2}}$ = $\sqrt {95}$
i.e. x=7, y=$\sqrt {95}$ and r=12
Given cos is positive and sin is negative, therefore $\theta$ lies in quadrant IV, hence-
$\sin\theta$ =$ -\frac{y}{r}$ = $- \frac{\sqrt {95}}{12}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{7}{12}$ (given)
$\tan\theta$ =$ -\frac{y}{x}$ =$- \frac{\sqrt {95}}{7}$
$\cot\theta$ =$ -\frac{x}{y}$ =$- \frac{7}{\sqrt {95}}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{12}{7}$
$\csc\theta$ =$ -\frac{r}{y}$ =$- \frac{12}{\sqrt {95}}$
