Question

${ \tan(1) }^{ - 1} + { \cos( - \frac{1}{2} ) }^{ - 1} + { \sin( \frac{ - 1}{2} ) }^{ - 1}$
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Answer 1

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${tan}^{ - 1} (1) + {cos}^{ - 1}( - \frac{1}{2} ) + {sin}^{ - 1} ( - \frac{1}{2} ) \\ let \: \: x = {tan}^{ - 1} 1 \implies \: tanx = 1 \\ x = \frac{\pi}{4} \\ since \: \: range \: \: of \: \: tan ^{ - 1} \: \: is \: \: ( - \frac{\pi}{2} \: \: \frac{\pi}{2} ) \\ hence \: \: the \: principal \: \: value \: \: is \: \: \frac{\pi}{4} \\ let \: \: y = {cos}^{ - 1} ( - \frac{1}{2} ) \\ cosy = - \frac{1}{2} \\ cosy = - cos( \frac{\pi}{3} ) \\ cosy = cos(\pi - \frac{\pi}{3} ) \\ y = \frac{2\pi}{3} \\ hence \: the \: principal \: value \: is \: \frac{2\pi}{3} \\ let \: z = {sin}^{ - 1} ( - \frac{1}{2} ) \\ sinz = - \frac{1}{2} \\ sinz = sin( - \frac{\pi}{6} ) \\ z = - \frac{\pi}{6} \\ now \: \: {tan}^{ - 1} (1) + {cos}^{ - 1} ( - \frac{1}{2} ) + {sin}^{ - 1} ( - \frac{1}{2} ) \\ = x + y + z \\ = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} \\ = \frac{3\pi + 8\pi - 2\pi}{12} \\ = \frac{9\pi}{12} \\ = \frac{3\pi}{4}$