# 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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