How do you evaluate $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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How do you evaluate $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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Answer 1 · 2016-03-03T04:58:29

$cos [arctan (- \frac{5}{12})] = \pm \frac{12}{13}$ $cos[arctan(-5/12)]=+-12/13$

Explanation:

$arctan (- \frac{5}{12})$ $arctan(-5/12)$ means an angle $θ$ $theta$ whose $tan θ = - \frac{5}{12}$ $tantheta=-5/12$ . As such we have to find $cos θ$ $costheta$ .

$cos θ = \frac{1}{sec θ} = \sqrt{\frac{1}{{sec}^{2} θ}} = \sqrt{\frac{1}{1 + {tan}^{2} θ}}$ $costheta=1/sectheta=sqrt(1/sec^2theta)=sqrt(1/(1+tan^2theta))$

Hence

$cos θ = \sqrt{\frac{1}{1 + {(- \frac{5}{12})}^{2}}} = \sqrt{\frac{1}{1 + \frac{25}{144}}}$ $costheta=sqrt(1/(1+(-5/12)^2))=sqrt(1/(1+25/144))$

= $\sqrt{\frac{1}{\frac{169}{144}}} = \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$ $sqrt(1/(169/144))=sqrt(144/169)=+-12/13$

Hence $cos [arctan (- \frac{5}{12})] = \pm \frac{12}{13}$ $cos[arctan(-5/12)]=+-12/13$

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How do you evaluate $cos [arctan (- \frac{5}{12})]$ $cos[arctan(-5/12)]$ ?

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Explanation:

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