How do you evaluate $a r c sec (sec (- \frac{π}{3}))$ $arcsec (sec(-pi/3))$ ?

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How do you evaluate $a r c sec (sec (- \frac{π}{3}))$ $arcsec (sec(-pi/3))$ ?

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Answer 1 · 2015-04-19T18:27:15

$\frac{π}{3}$ $pi/3$

First $sec (- \frac{π}{3})$ $sec(-pi/3)$ would be equal to $sec (\frac{π}{3})$ $sec(pi/3)$ = 2. ( Angle - $\frac{π}{3}$ $pi/3$ lies in the fourth quadrant and because cosine is positive there sec would also be positive).

Now the problem is reduced to evaluating arcsec2. Since $sec (\frac{π}{3})$ $sec(pi/3)$ is 2, hence arcsec2 would be equal to $\frac{π}{3}$ $pi/3$

Answer 2 · 2015-04-19T18:30:55

$θ = a r c sec (sec (- \frac{π}{3}))$ $theta=arcsec(sec(-pi/3))$

$sec θ = sec (- \frac{π}{3})$ $sectheta=sec(-pi/3)$

$:.theta=-pi/3$

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How do you evaluate $a r c sec (sec (- \frac{π}{3}))$ $arcsec (sec(-pi/3))$ ?

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