Question 19 of 42intermediate🔧 ApplyMCQ1 mark

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

Exercise: EXERCISE 2.2 | Q: 15 | (Chapter: 13)
For More Understanding

Explanation

We need to evaluate the expression tan⁡−13−cot⁡−1(−3)\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}).

Step 1: Evaluate tan⁡−13\tan^{-1}\sqrt{3} Let tan⁡−13=x\tan^{-1}\sqrt{3} = x. Then tan⁡x=3\tan x = \sqrt{3}. Since tan⁡(π/3)=3\tan(\pi/3) = \sqrt{3} and π/3\pi/3 lies in the principal value branch of tan⁡−1\tan^{-1}, which is (−π/2,π/2)(-\pi/2, \pi/2), we have: tan⁡−13=π/3\tan^{-1}\sqrt{3} = \pi/3.

Step 2: Evaluate cot⁡−1(−3)\cot^{-1}(-\sqrt{3}) Let cot⁡−1(−3)=y\cot^{-1}(-\sqrt{3}) = y. Then cot⁡y=−3\cot y = -\sqrt{3}. We know cot⁡(π/6)=3\cot(\pi/6) = \sqrt{3}, so −3=−cot⁡(π/6)=cot⁡(π−π/6)=cot⁡(5π/6)-\sqrt{3} = -\cot(\pi/6) = \cot(\pi - \pi/6) = \cot(5\pi/6). The principal value branch of cot⁡−1\cot^{-1} is (0,π)(0, \pi), and 5π/65\pi/6 lies within this interval. Thus, cot⁡−1(−3)=5π/6\cot^{-1}(-\sqrt{3}) = 5\pi/6.

Step 3: Calculate the final value Substituting the values found: tan⁡−13−cot⁡−1(−3)=π3−5π6\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6} =2π6−5π6=−3π6=−π2= \frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}.

Thus, the correct option is (B).

Solution Steps

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

  2. Step 2: Find the principal value of cot⁻¹(-3\sqrt{3}). Let cot⁻¹(-3\sqrt{3}) = y. Then cot y = -3\sqrt{3} = cot(π - π/6) = cot(5π/6). Since 5π/6 ∈ (0, π), y = 5π/6.

  3. Step 3: Subtract the values: π/3 - 5π/6 = 2π/6 - 5π/6 = -3π/6 = -π/2.

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