$T h e v a l u e o f \sin^{- 1} (\cos 2) - \cos^{- 1} (\sin 2) + \tan^{- 1} (\cot 4) - \cot^{- 1} (\tan 4) + \sec^{- 1} (cosec 6) - {cosec}^{- 1} (\sec 6)$

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$3 π$

$5 π - 16$

$8 - 3 π$

answer is D.

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Detailed Solution

$\sin^{- 1} (\cos 2) - \cos^{- 1} (\sin 2) + \tan^{- 1} (\cot 4) - \cot^{- 1} (\tan 4) + \sec^{- 1} (cosec 6) - {cosec}^{- 1} (\sec 6)$ $= \frac{π}{2} - c o s^{- 1} (\cos 2) - (\frac{π}{2} - \sin^{- 1} (\sin 2)) + \frac{π}{2} - c o t^{- 1} (\cot 4) - (\frac{π}{2} - \tan^{- 1} (\tan 4)) + \frac{π}{2} - c o \sec^{- 1} (cosec 6) - (\frac{π}{2} - \sec^{- 1} (\sec 6)) = \sin^{- 1} (\sin 2) - c o s^{- 1} (\cos 2) + \tan^{- 1} (\tan 4) - c o t^{- 1} (\cot 4) + \sec^{- 1} (\sec 6) - c o \sec^{- 1} (cosec 6)$ $\begin{array}{l} = (π - 2) - 2 + (4 - π) - (4 - π) + (2 π - 6) - (6 - 2 π) \\ = π - 4 + 4 π - 12 = 5 π - 16 \end{array}$