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 $cos(theta)=-1/3$

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

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