How do you find the value of #tan theta# given #sec theta=-9/4# and in quadrant III?

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How do you find the value of #tan theta# given #sec theta=-9/4# and in quadrant III?

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Answer 1 · 2016-12-17T05:11:37

#sqrt65/4#

Explanation:

#cos t = 1/(sec t )= - 4/9#
#sin^2 t = 1 - cos^2 t = 1 - (4/9)^2 = (81 - 16)/81 = 65/81# -->
#sin t = +- sqrt65/9#
Since t is in Quadrant III, then sin t is negative.
#tan t = sint/(cos t) = (- sqrt65/9)(-9/4) = sqrt65/4#

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How do you find the value of #tan theta# given #sec theta=-9/4# and in quadrant III?

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