Question:
\( \left[ \tan^{-1}\sqrt{3} - \sec^{-1}(-2) \right] \) is equal to :
\( \left[ \tan^{-1}\sqrt{3} - \sec^{-1}(-2) \right] \) is equal to :
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Principal values: tan inverse of root 3 is pi/3, sec inverse of -2 is 2pi/3.
Updated On: Oct 22, 2026
- \( \pi \)
- \( -\dfrac{\pi}{3} \)
- \( \dfrac{\pi}{3} \)
- \( \dfrac{2\pi}{3} \)
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The Correct Option is B
Solution and Explanation
Step 1: Recall the principal value ranges:
The principal range of \( \tan^{-1} \) is \( \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \). The principal range of \( \sec^{-1} \) is \( [0, \pi] - \left\{\dfrac{\pi}{2}\right\} \).
Step 2: Find the first term:
Since \( \tan\dfrac{\pi}{3} = \sqrt{3} \) and \( \dfrac{\pi}{3} \) lies in the range, \( \tan^{-1}\sqrt{3} = \dfrac{\pi}{3} \).
Step 3: Find the second term:
Use \( \sec^{-1}(-x) = \pi - \sec^{-1}x \) for \( |x| \geq 1 \). Then \( \sec^{-1}(-2) = \pi - \sec^{-1}2 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} \).
Step 4: Subtract:
\[ \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3} \]
Step 5: Check the other options:
Option (A) \( \pi \) would need the terms to add, not subtract. Option (C) \( \dfrac{\pi}{3} \) forgets the second term. Option (D) \( \dfrac{2\pi}{3} \) is the value of \( \sec^{-1}(-2) \) alone.
Final Answer:
The value is \( -\dfrac{\pi}{3} \), which is option (B). \[ \boxed{-\dfrac{\pi}{3}} \]
The principal range of \( \tan^{-1} \) is \( \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \). The principal range of \( \sec^{-1} \) is \( [0, \pi] - \left\{\dfrac{\pi}{2}\right\} \).
Step 2: Find the first term:
Since \( \tan\dfrac{\pi}{3} = \sqrt{3} \) and \( \dfrac{\pi}{3} \) lies in the range, \( \tan^{-1}\sqrt{3} = \dfrac{\pi}{3} \).
Step 3: Find the second term:
Use \( \sec^{-1}(-x) = \pi - \sec^{-1}x \) for \( |x| \geq 1 \). Then \( \sec^{-1}(-2) = \pi - \sec^{-1}2 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} \).
Step 4: Subtract:
\[ \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3} \]
Step 5: Check the other options:
Option (A) \( \pi \) would need the terms to add, not subtract. Option (C) \( \dfrac{\pi}{3} \) forgets the second term. Option (D) \( \dfrac{2\pi}{3} \) is the value of \( \sec^{-1}(-2) \) alone.
Final Answer:
The value is \( -\dfrac{\pi}{3} \), which is option (B). \[ \boxed{-\dfrac{\pi}{3}} \]
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