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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using the principal values, evaluate the following \[ tan^{-1}\frac{1}{\sqrt{3}}+sin^{-1} \bigg( -\frac{1}{2} \bigg) \]

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\[\frac{\pi}{6}-\frac{\pi}{6}=0\]

 

We know that \(tan\frac{\pi}{6}=\frac{1}{\sqrt3}\) and \(\frac{\pi}{6}\:\in\:(-\frac{\pi}{2},\frac{\pi}{2})\)

which is the principal interval of tan

Therefore,\(tan^{-1}\frac{1}{\sqrt3}=tan^{-1}tan\frac{\pi}{6}=\frac{\pi}{6}\)

similarly  we know that  \(sin(-\frac{\pi}{6})=-\frac{1}{2}\)  and

\(-\frac{\pi}{6}\in\:[-\frac{\pi}{2},\frac{\pi}{2}]\) which is the principal interval of sin

Therefore, \(sin^{-1}(-\frac{1}{2})=sin^{-1}sin(-\frac{\pi}{6})=-\frac{\pi}{6}\)

\(\Rightarrow\:tan^{-1}\frac{1}{\sqrt3}+sin^{-1}-\frac{1}{2}=\frac{\pi}{6}-\frac{\pi}{6}=0\)
answered Feb 13, 2013 by rvidyagovindarajan_1
edited Mar 19, 2013 by rvidyagovindarajan_1
 

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