# Write the first trigonometric function in terms of the second for \theta i

Write the first trigonometric function in terms of the second for $$\displaystyle\theta$$ in the given quadrant
$$\displaystyle{\tan{\theta}},\ {\cos{\theta}};\ \theta\ \text{ in Quadrant II}$$

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i1ziZ
Since $$\displaystyle{\tan{\theta}}={\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}$$, we need to write $$\displaystyle{\sin{\theta}}$$ in terms of $$\displaystyle{\cos{\theta}}$$.
$$\displaystyle{\sin{\theta}}=\pm\sqrt{{{1}-{{\cos}^{{2}}\theta}}}$$
and since $$\displaystyle{\sin{\theta}}$$ is negative in Quadrant III, the negative sign applies here. Thus
$$\displaystyle{\tan{\theta}}={\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={\frac{{-\sqrt{{{1}-{{\cos}^{{2}}\theta}}}}}{{{\cos{\theta}}}}}$$
Result: $$\displaystyle{\frac{{-\sqrt{{{1}-{{\cos}^{{2}}\theta}}}}}{{{\cos{\theta}}}}}$$