How do you find the exact value of #tan[arc cos(-1/3)]#?

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Answer 1 · 2015-07-28T10:17:52

You use the trigonometric Identity #tan(theta)=sqrt((1/cos^2(theta)-1))#

Result : #tan[arccos(-1/3)]=color(blue)(2sqrt(2))#

Start by letting #arccos(-1/3)# to be an angle #theta#

#=>arccos(-1/3)=theta#

#=>cos(theta)=-1/3#

This means that we are now looking for #tan(theta)#

Next, use the identity : #cos^2(theta)+sin^2(theta)=1#

Divide all both sides by #cos^2(theta)# to have,

#1+tan^2(theta)=1/cos^2(theta)#

#=>tan^2(theta)=1/cos^2(theta)-1#

#=>tan(theta)=sqrt((1/cos^2(theta)-1))#

Recall, we said earlier that #cos(theta)=-1/3#

#=>tan(theta)=sqrt(1/(-1/3)^2-1)=sqrt(1/(1/9)-1)=sqrt(9-1)=sqrt(8)=sqrt(4xx2)=sqrt(4)xxsqrt(2)=color(blue)(2sqrt(2))#