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Question 68 of 71

Q.Prove that tan⁡−1211+tan⁡−1724=tan⁡−112\tan^{-1}\dfrac{2}{11}+\tan^{-1}\dfrac{7}{24}=\tan^{-1}\dfrac12

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2025Subjective· 3mImportance★★★★★
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Applies the addition formula tan⁡−1A+tan⁡−1B=tan⁡−1 ⁣(A+B1−AB)\tan^{-1}A+\tan^{-1}B=\tan^{-1}\!\left(\dfrac{A+B}{1-AB}\right) (valid here since AB<1AB<1) and simplifies the resulting fraction to 12\dfrac12.

  1. Let A=211A=\dfrac{2}{11} and B=724B=\dfrac{7}{24}; both are positive.
  2. Compute AB=211⋅724=14264=7132AB=\dfrac{2}{11}\cdot\dfrac{7}{24}=\dfrac{14}{264}=\dfrac{7}{132}. Since AB=7132<1AB=\dfrac{7}{132}<1, the addition formula tan⁡−1A+tan⁡−1B=tan⁡−1 ⁣(A+B1−AB)\tan^{-1}A+\tan^{-1}B=\tan^{-1}\!\left(\dfrac{A+B}{1-AB}\right) applies directly (no adjustment by π\pi is needed), and the sum stays within the principal range because both A,B>0A,B>0 are small.
  3. Compute A+B=211+724A+B=\dfrac{2}{11}+\dfrac{7}{24}. Using LCM 264264: 211=48264\dfrac{2}{11}=\dfrac{48}{264} and 724=77264\dfrac{7}{24}=\dfrac{77}{264}, so A+B=48+77264=125264A+B=\dfrac{48+77}{264}=\dfrac{125}{264}. …

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