How do you find the exact value of tan[arccos(−13)]?

1 Answer
Antoine
Jul 28, 2015

You use the trigonometric Identity tan(θ)=√(1cos2(θ)−1)

Result : tan[arccos(−13)]=2√2

Explanation:

Start by letting arccos(−13) to be an angle θ

⇒arccos(−13)=θ

⇒cos(θ)=−13

This means that we are now looking for tan(θ)

Next, use the identity : cos2(θ)+sin2(θ)=1

Divide all both sides by cos2(θ) to have,

1+tan2(θ)=1cos2(θ)

⇒tan2(θ)=1cos2(θ)−1

⇒tan(θ)=√(1cos2(θ)−1)

Recall, we said earlier that cos(θ)=−13

⇒tan(θ)=  ⎷1(−13)2−1=√119−1=√9−1=√8=√4×2=√4×√2=2√2

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