Let 0 be an angle in quadrant II such that sin0 =2/3. Find the exact values of sec 0 and cot 0?

Question

James Adkins · Accepted Answer

#sec theta = -3/sqrt5#

#cot theta = - sqrt5/2# #sec theta = 1/cos theta# #cot theta = cos theta/sin theta# We know that #sin^2 theta + cos^2 theta = 1# From which we get #cos theta = +-sqrt(1-sin^2 theta)#. Since #theta# is in the second quadrant, that means #cos theta# is negative. In other words, #cos theta = color(red)-sqrt(1-sin^2 theta)# Substitute #sin theta = 2/3# into the equation to get : #cos theta = -sqrt(1-4/9) = - sqrt5/3# Therefore, #color(blue)(sec theta) = 1/(-sqrt5/3) = color(blue)(- 3/sqrt5)# and #color(blue)(cot theta) = (-sqrt5/3)/(2/3) = color(blue)(-sqrt 5/2)#.

Grace Ashcraft · Answer

undefined Given that $ \sin(\theta) = \frac{2}{3} $, and $ \theta $ is an angle in quadrant II, we can find the exact values of $ \sec(\theta) $ and $ \cot(\theta) $ using trigonometric identities.

We know that $ \sin(\theta) = \frac{2}{3} $. Since $ \theta $ is in quadrant II, $ \cos(\theta) $ will be negative. Using the Pythagorean identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $, we can find $ \cos(\theta) $:
$$ \cos^2(\theta) = 1 - \sin^2(\theta) $$
$$ \cos^2(\theta) = 1 - \left(\frac{2}{3}\right)^2 $$
$$ \cos^2(\theta) = 1 - \frac{4}{9} $$
$$ \cos^2(\theta) = \frac{9}{9} - \frac{4}{9} $$
$$ \cos^2(\theta) = \frac{5}{9} $$
$$ \cos(\theta) = -\frac{\sqrt{5}}{3} $$

Now, we can find $ \sec(\theta) $ and $ \cot(\theta) $:
$$ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{\sqrt{5}}{3}} = -\frac{3}{\sqrt{5}} = -\frac{3\sqrt{5}}{5} $$

$$ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{-\frac{\sqrt{5}}{3}}{\frac{2}{3}} = -\frac{\sqrt{5}}{2} $$

Therefore, the exact values of $ \sec(\theta) $ and $ \cot(\theta) $ are $ -\frac{3\sqrt{5}}{5} $ and $ -\frac{\sqrt{5}}{2} $, respectively.