Given #sectheta=sqrt2, sintheta=-sqrt2/2# to find the remaining trigonometric function?

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ciumes Share
Apr 13, 2018

If #sec theta=sqrt2# then draw a right triangle in standard position with #x=1#, #r=sqrt2#. Now, with #sin theta# negative, it will be in QIII or QIV. SInce #x# is positive, it is in QIV. Now you know #y=-1/sqrt2#
#sin theta=y/r = (-1/sqrt2)/sqrt2=-1/2#
#cos theta=1/sec theta=1/sqrt2#
#tan theta = y/x = -1#
#cot theta = x/y = -1#
#csc theta = r/y = -2#

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