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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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Toolbox:
• $sin\large\frac{\pi}{2}=1$
• $-tan\theta=tan(-\theta)$
• $\large\frac{1}{\sqrt{3}}=cot\large\frac{\pi}{3}$
• $tan^{-1}1=\large\frac{\pi}{4}$
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
edited Mar 19, 2013