Question 16 of 42intermediate🔧 ApplyMCQ1 mark

tan⁻¹3\sqrt{3} - sec⁻¹(-2) is equal to

Exercise: EXERCISE 2.1 | Q: 14 | (Chapter: 10)
For More Understanding

Explanation

Option B is correct. The principal value branch of sec⁻¹ is given in the context as [0, π] - {π/2}. For sec⁻¹(-2), we find θ in this interval with sec θ = -2, i.e., cos θ = -1/2, which gives θ = 2π/3. For tan⁻¹3\sqrt{3}, the principal value branch is (-π/2, π/2) (implied by the discussion on restricting the domain of the tangent function). Since tan(π/3) = 3\sqrt{3} and π/3 lies in (-π/2, π/2), we have tan⁻¹3\sqrt{3} = π/3. The difference π/3 - 2π/3 = -π/3, matching option B.

Solution Steps

  1. Step 1: Find tan⁻¹3\sqrt{3}. tan(π/3) = 3\sqrt{3} and π/3 ∈ (-π/2, π/2), the principal value branch of tan⁻¹. Hence tan⁻¹3\sqrt{3} = π/3.

  2. Step 2: Find sec⁻¹(-2). The principal value branch of sec⁻¹ is [0, π] - {π/2}. Solve sec θ = -2 ⇒ cos θ = -1/2. In [0, π], cos θ = -1/2 at θ = 2π/3. Thus sec⁻¹(-2) = 2π/3.

  3. Step 3: Compute tan⁻¹3\sqrt{3} - sec⁻¹(-2) = π/3 - 2π/3 = -π/3. Therefore, the correct option is (B).

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