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Q.Prove that 2tan⁡−112+tan⁡−117=tan⁡−131172\tan^{-1}\dfrac12+\tan^{-1}\dfrac17=\tan^{-1}\dfrac{31}{17}

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Converts the double-angle term using 2tan⁡−1x=tan⁡−1(2x1−x2)2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^2}\right), then combines with the remaining term via the tangent addition formula.

  1. First simplify 2tan⁡−1122\tan^{-1}\dfrac12 using 2tan⁡−1x=tan⁡−1(2x1−x2)2\tan^{-1}x=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right), valid here since x=12x=\dfrac12 gives x2=14<1x^2=\dfrac14<1.
  2. With x=12x=\dfrac12: 2x1−x2=2(1/2)1−1/4=13/4=43\dfrac{2x}{1-x^2}=\dfrac{2(1/2)}{1-1/4}=\dfrac{1}{3/4}=\dfrac43. So 2tan⁡−112=tan⁡−1432\tan^{-1}\dfrac12=\tan^{-1}\dfrac43.
  3. The LHS becomes tan⁡−143+tan⁡−117\tan^{-1}\dfrac43+\tan^{-1}\dfrac17. Use 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 when ab<1ab<1.
  4. Here a=43, b=17a=\dfrac43,\ b=\dfrac17: ab=421<1ab=\dfrac{4}{21}<1, so the formula applies directly. …

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