<s>[INST] Find the value of $\sin^{-1}(\cos 2)-\cos^{-1}(\sin 2) +\tan^{-1}(\cot 4) -\cot^{-1}(\tan 4)+\sec^{-1}(\csc 6)-\csc^{-1}(\sec 6)$ The given expression simplifies to
$$\sin^{-1}(\sin 2)-\cos^{-1}(\cos 2)+\tan^{-1}(\tan 4)-\cot^{-1}(\cot 4)+\sec^{-1}(\sec 6)-\csc^{-1} (\csc 6)$$
$$=(\pi-2)-2+(4-\pi)-(2\pi-4)+(2\pi-6)-(2\pi-6)$$
$$=-2\pi+4$$
But the given answer is $5\pi-16$. I rechecked all the the principle branches, and they all seem to be right. Where did use the wrong value?
 [/INST] I don't get your simplification. How did $\sin^{-1}(\cos 2) - \cos^{-1}(\sin 2)$ become $\sin^{-1}(\sin 2) - \cos^{-1}(\cos 2)$? I think the simplification part is off.

We must identify which quadrant each of these lie in. For example, $2$ lies in the second quadrant, because $\pi > 2 > \frac \pi 2$. $4$ lies in the third quadrant, and $6$ lies in the fourth quadrant.
With this in mind, $\sin^{-1}(\cos 2) = \frac{\pi}{2} - 2$. Similarly, $\cos^{-1}(\sin 2) = 2 - \frac \pi 2$, because the inverse cosine function is restricted to $(-\frac{\pi}{2},\frac{\pi}{2}]$.

*

*$\tan^{-1}(\cot 4) = \frac{3 \pi}{2} - 4$.


*$\cot^{-1} (\tan 4) = \frac{3 \pi }{2} - 4$.


*$\sec^{-1}(\csc 6) = \frac{5 \pi}{2} - 6$.


*$\csc^{-1}(\sec 6) = 6 - \frac{3 \pi}{2}$.
(You can convince yourself of these and finish the problem).

With the simplified expression :

*

*$\sin^{-1}(\sin 2) = \pi- 2$


*$\cos^{-1}(\cos 2) = 2$


*$\tan^{-1}(\tan 4) = 4-\pi$.


*$\color{green}{\cot^{-1}(\cot 4) = 4-\pi}$. Explanation : the principal region for $\cot^{-1}$ is $[-\frac \pi 2, \frac \pi 2]$, and $\cot 4$ is positive because $4$ is in the third quadrant. Thus, $\cot^{-1}(\cot 4)$ lies in the first quadrant, so the angle is positive, hence $4-\pi$ (and not $\pi -4$, which is a negative angle).


*$\sec^{-1}(\sec 6) = 2\pi - 6$.


*$\color{green}{\csc^{-1}(\csc 6) = 6-2\pi}$. Explanation : the principal region of $\csc^{-1}$ is $[-\frac {\pi}{2} , \frac{\pi}{2}]$ , and  $6$ belongs in the fourth quadrant, so is already in the prescribed region, as long as we subtract $2\pi$, which gives $6 - 2\pi$ (and since the cosecants of both these angles are negative, this makes sense).
Now evaluating gives the right answer.
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