Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

We simplify each inverse trigonometric term using periodicity and range considerations.

${\sin }^{-1}(\sin (\frac{33\pi }{7}))={\sin }^{-1}(\sin (\frac{33\pi }{7}-4\pi ))={\sin }^{-1}(\sin (\frac{5\pi }{7}))={\sin }^{-1}(\sin (\pi -\frac{2\pi }{7}))=\frac{2\pi }{7}$

${\cos }^{-1}(\cos (\frac{46\pi }{7}))={\cos }^{-1}(\cos (\frac{46\pi }{7}-6\pi ))={\cos }^{-1}(\cos (\frac{4\pi }{7}))=\frac{4\pi }{7}$

${\tan }^{-1}(-\tan (\frac{13\pi }{8}))={\tan }^{-1}(\tan (-\frac{13\pi }{8}))={\tan }^{-1}(\tan (-\frac{13\pi }{8}+2\pi ))={\tan }^{-1}(\tan (\frac{3\pi }{8}))=\frac{3\pi }{8}$

${\cot }^{-1}(\cot (-\frac{19\pi }{8}))={\cot }^{-1}(\cot (-\frac{19\pi }{8}+3\pi ))={\cot }^{-1}(\cot (\frac{5\pi }{8}))=\frac{5\pi }{8}$

Adding these, we get $\frac{2\pi }{7}+\frac{4\pi }{7}+\frac{3\pi }{8}+\frac{5\pi }{8}=\frac{6\pi }{7}+\pi =\frac{13\pi }{7}$.

Thus, $a=13$ and $b=7$, so $a+b=13+7=20$.

Final Answer: The final answer is $\boxed{20}$