How do you simplify the expression $sec (arccos (- \frac{3}{5}))$ $sec(arccos (-3/5))$ ?

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How do you simplify the expression $sec (arccos (- \frac{3}{5}))$ $sec(arccos (-3/5))$ ?

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Answer 1 · 2016-07-04T14:13:16

$sec (arccos (- \frac{3}{5})) = - \frac{5}{3}$ $sec(arccos(-3/5))=-5/3$

Explanation:

Let $arccos (- \frac{3}{5}) = θ$ $arccos(-3/5)=theta$

Therefore $cos θ = - \frac{3}{5}$ $costheta=-3/5$

and hence $sec (arccos (- \frac{3}{5})) = sec θ$ $sec(arccos(-3/5))=sectheta$

= $\frac{1}{cos θ} = \frac{1}{- \frac{3}{5}}$ $1/costheta=1/(-3/5)$

= $1 \times - \frac{5}{3} = - \frac{5}{3}$ $1xx-5/3=-5/3$

Answer 2 · 2016-07-04T16:28:05

$- \frac{5}{3}$ $-5/3$

Explanation:

Use cos arc cos x = x.

Here, $sec a r c cos (- \frac{3}{5}) = \frac{1}{cos a r c cos (- \frac{3}{5})} = \frac{1}{- \frac{3}{5}} = - \frac{5}{3}$ $sec arc cos (-3/5)=1/(cos arc cos (-3/5))=1/(-3/5)=-5/3$

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