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How do you find the five remaining trigonometric function satisfying #costheta=-2/5#, #sintheta>0#?

1 Answer

Aug 4, 2018

#sin theta= sqrt 21/5 , csc theta=5/sqrt 21 ,tan theta= - sqrt 21/2#
#cot theta= -2/sqrt 21 , sec theta=-5/2#

Explanation:

In right triangle #cos theta = a/h =-2 /5 :.a=-2 , h=5 #

#:.o=sqrt(5^2-2^2)= sqrt 21 ; a,,o, h # are adjacent side, opposite

side and hypotenuse. #sin theta= o/h= sqrt 21/5# ,

#tan theta= o/a= - sqrt 21/2, csc theta= h/o= 5/sqrt 21# ,

#sec theta= h/a= - 5/2, cot theta= a/o= -2/sqrt 21# [Ans]

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