Find the value of $sin^{-1}({-\sqrt{3}\over 2})$ + $cos^{-1}(cos({7\pi\over 6}))$.

Solution :

$sin^{-1}({-\sqrt{3}\over 2})$ = – $sin^{-1}({\sqrt{3}\over 2})$ = $-\pi\over 3$

$cos^{-1}(cos({7\pi\over 6}))$ = $cos^{-1}(cos({2\pi – {5\pi\over 6}}))$ = $cos^{-1}(cos({5\pi\over 6}))$ = $5\pi\over 6$

Hence $sin^{-1}({-\sqrt{3}\over 2})$ + $cos^{-1}(cos({7\pi\over 6}))$ = $-\pi\over 3$ + $5\pi\over 6$ = $\pi\over 2$

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