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Home  >>  CBSE XII  >>  Math
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Find the value of $tan^{-1}\frac{-1}{\sqrt 3}+cot^{-1}\frac{1}{\sqrt 3}-tan^{-1}sin\frac{\pi}{2}$

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  • \(sin\large\frac{\pi}{2}=1\)
  • \(-tan\theta=tan(-\theta)\)
  • \(\large\frac{1}{\sqrt{3}}=cot\large\frac{\pi}{3}\)
  • \(tan^{-1}1=\large\frac{\pi}{4}\)
By taking \(\theta=\large\frac{\pi}{6}\)
\(-\large\frac{1}{\sqrt{3}}=-tan\large\frac{\pi}{6}=tan(-\large\frac{\pi}{6})\)
\(tan^{-1}(-\large\frac{1}{\sqrt{3}})=tan^{-1}tan(-\large\frac{\pi}{6})= -\large\frac{\pi}{6}\)
\(cot^{-1}\large\frac{1}{\sqrt{3}}=cot^{-1}cot\large\frac{\pi}{3}=\large\frac{\pi}{3}\)
\(tan^{-1}sin\large\frac{\pi}{2}=tan^{-1}1=\large\frac{\pi}{2}\)
\(tan^{-1}(-\large\frac{1}{\sqrt{3}})+ cot^{-1}\large\frac{1}{\sqrt{3}}-tan^{-1}sin\large\frac{\pi}{2}=-\large\frac{\pi}{6}+\large\frac{\pi}{3}-\large\frac{\pi}{4}\)
\( = -\large\frac{\pi}{12}\)

 

answered Mar 2, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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