Reciprocal trig ratios; inverse trig functions practice questions.

Given that sec θ = 3, where 0 < θ < π/2, determine the exact value of tan θ.

• 2
• √10
• 2√2
• 8

Answer: 2√2

Using 1 + tan²θ = sec²θ gives tan²θ = 9 − 1 = 8, and since θ is in the first quadrant tan θ = √8 = 2√2.

State the exact value of csc(π/6).

• 1/2
• 2
• √2
• 2/√3

Answer: 2

csc θ = 1/sin θ, and sin(π/6) = 1/2, so csc(π/6) = 2. A common error is to compute 1/cos(π/6) instead.

State the range of the function f(x) = arccos x.

• [−π/2, π/2]
• (−π/2, π/2)
• [0, 2π]
• [0, π]

Answer: [0, π]

arccos is defined by restricting cos to [0, π], where it is one-to-one, so the range of arccos is [0, π]. The interval [−π/2, π/2] is the range of arcsin.

Determine the exact value of arctan(√3).

• π/3
• π/6
• π/4
• 2π/3

Answer: π/3

tan(π/3) = √3 and π/3 lies in the range (−π/2, π/2) of arctan, so arctan(√3) = π/3. Choosing π/6 confuses tan with its reciprocal, since tan(π/6) = 1/√3.

Which of the following identities holds for all θ where the functions are defined?

• 1 + csc²θ = cot²θ
• 1 + cot²θ = csc²θ
• cot²θ − 1 = csc²θ
• 1 + sec²θ = tan²θ

Answer: 1 + cot²θ = csc²θ

Dividing sin²θ + cos²θ = 1 by sin²θ gives 1 + cot²θ = csc²θ. The other options swap the roles of the squared functions.

State the exact value of cot(π/3).

• √3
• 2/√3
• 1/2
• 1/√3

Answer: 1/√3

cot θ = 1/tan θ and tan(π/3) = √3, so cot(π/3) = 1/√3 = √3/3. Answering √3 forgets to take the reciprocal.