\(\rm cot^{-1}{1\over3} - 2 \tan^{-1}{2\over3} = ?\)

  1. 1
  2. -1
  3. \(\rm \cot^{-1}{41\over3}\)
  4. \(\rm \tan^{-1}{41\over3}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \cot^{-1}{41\over3}\)
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Detailed Solution

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Concept:

  • \(\rm \tan^{-1}x + \tan^{-1}y=\tan^{-1}{x+y\over 1-xy}\)
  • \(\rm cot^{-1}{x} = {\pi\over2}- \tan^{-1}{x}\)
  • \(\rm 2tan^{-1}\ x =tan^{-1} ({\frac {2x}{1\ -\ x^2}})\)


Calculation:

S = \(\rm cot^{-1}{1\over3} - 2 \tan^{-1}{2\over3} \)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{2\times \frac{2}{3}}\over1-{(\frac{2}{3})^2}}\)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{\frac{4}{3}}\over1-{\frac{4}{9}}}\)

S = \(\rm {\pi\over2}-\tan^{-1}{1\over3} - \tan^{-1}{\frac{12}{5}}\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{12\over5}+\tan^{-1}{1\over3}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{{12\over5}+{1\over3}\over1-{12\over5}\times{1\over3}}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{41\over3}\right]\)

S = \(\rm \cot^{-1}{41\over3}\)

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