How do you determine #tantheta# given #cottheta=-sqrt5/2,pi/2<theta<pi#?

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Answer 1 · 2017-01-24T05:45:43

#tantheta=-2/sqrt5#; #sintheta=-2/3#; #csctheta=-3/2#;#costheta=-sqrt5/3#;#sectheta=-3/sqrt5#

Since #tantheta=1/cottheta#, you get:

#tantheta=-2/sqrt5#

Since #sintheta=+-tantheta/sqrt(1+tan^2theta)# and sine is positive in the second quadrant:

#sintheta=+(-2/sqrt5)/sqrt(1+(-2/sqrt5)^2#

#=(-2/sqrt5)/sqrt(1+4/5)=(-2/sqrt5)/sqrt(9/5)=-2/cancelsqrt5*cancelsqrt5/3=-2/3#

Then #csctheta=1/sintheta=-3/2#

#costheta=+-sqrt(1-sin^2theta)#

In the second quadrant cosine is negative, then

#costheta=-sqrt(1-(-2/3)^2)=-sqrt(1-4/9)=-sqrt(5/9)=-sqrt5/3#

and #sectheta=1/costheta=-3/sqrt5#