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Question 69 of 71

Q.If x<0x<0, then tan⁡−1(1x)\tan^{-1}\left(\dfrac1x\right) is equal to :

(a) −π+cot⁡−1(x)-\pi+\cot^{-1}(x)
(b) tan⁡−1(x)\tan^{-1}(x)
(c) −π+tan⁡−1x-\pi+\tan^{-1}x
(d) cot⁡−1(x)\cot^{-1}(x)
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026MCQ· 1mImportance★★★★★
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Derives the identity relating tan⁡−1(1/x)\tan^{-1}(1/x) and cot⁡−1x\cot^{-1}x for negative xx by comparing ranges, then confirms with a numeric check.

  1. For x>0x>0, the standard identity is tan⁡−1(1x)=cot⁡−1x\tan^{-1}\left(\dfrac1x\right)=\cot^{-1}x, both lying in (0,π2)\left(0,\dfrac\pi2\right).
  2. For x<0x<0, 1x<0\dfrac1x<0 too, so tan⁡−1(1x)∈(−π2,0)\tan^{-1}\left(\dfrac1x\right)\in\left(-\dfrac\pi2,0\right) (using the principal range of tan⁡−1\tan^{-1}). …

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