Given cos β = - 4/9 with β in the third quadrant, what is the exact value of tan β/2?

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Answer 1 · 2018-03-29T19:32:10

#tan(beta/2)=-sqrt(13/5)#

We have,

#cosbeta=-4/9,#where,#beta in 3rd# quadrant

#=>beta in [pi,(3pi)/2]=>(beta)/2 in [pi/2,(3pi)/4]#

i.e.2nd quadrant #=>tan(beta/2)<0#

#tan(beta/2)=-sqrt((1-cosbeta)/(1+cosbeta)#

#=-sqrt((1+(4/9))/(1-(4/9)#

#=-sqrt((9+4)/(9-4)#

#=-sqrt(13/5)#