The value of $\sin^{- 1} (\cos 2) - \cos^{- 1} (\sin 2) + \tan^{- 1} (\cot (4)) - \cot^{- 1} (\tan 4) + \sec^{- 1} (cosec 6) - {cosec}^{- 1} (\sec 6)$ is

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$3 π$

$8 - 3 π$

$5 π - 16$

answer is D.

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Detailed Solution

$\begin{array}{l} \sin^{- 1} (\cos 2) = \frac{π}{2} - \cos^{- 1} (\cos 2) = \frac{π}{2} - 2 \cos^{- 1} (\sin 2) = \frac{π}{2} - \sin^{- 1} (\sin 2) = \frac{π}{2} - \sin^{- 1} (\sin (π - 2)) = \frac{π}{2} - π + 2 = 2 - \frac{π}{2} \\ \tan^{- 1} (\cot 4) = \frac{π}{2} - \cot^{- 1} (\cot (- 4)) = \frac{π}{2} - \cot^{- 1} (\cot (4 - π)) = \frac{π}{2} - (4 - π) = \frac{3 π}{2} - 4 \\ \cot^{- 1} (\tan 4) = \frac{π}{2} - \tan^{- 1} (\tan (4)) = \frac{π}{2} - \tan^{- 1} (\tan (4 - π)) \frac{π}{2} - (4 - π) = \frac{3 π}{2} - 4 \\ \sec^{- 1} (cosec 6) = \frac{π}{2} - cosec (cosec 6) = \frac{π}{2} - {cosec}^{- 1} (cosec (6 - 2 π)) \\ = \frac{π}{2} - (6 - 2 π) = \frac{5 π}{2} - 6 cosec (\sec 6) = \frac{π}{2} - \sec^{- 1} (\sec 6) = \frac{π}{2} - \sec^{- 1} (\sec (2 π - 6)) \\ = \frac{π}{2} - (2 π - 6) = \frac{- 3 π}{2} + 6 \\ \sin^{- 1} (\cos 2) - \cos^{- 1} (\sin 2) + \tan^{- 1} (\cot 4) - \cot^{- 1} (\tan 4) + \sec^{- 1} (cosec 6) - {cosec}^{- 1} (\sec 6) \\ = \frac{π}{2} - 2 - 2 + \frac{π}{2} + \frac{3 π}{2} - 4 - \frac{3 π}{2} + 4 + \frac{5 π}{2} - 6 + \frac{3 π}{2} - 6 = 5 π - 16 \end{array}$