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Q)

Evaluate : $\cos\; \large \bigg( \frac{\pi}{2} -$$ \cos^{-1}\; \large\bigg (-\frac{\sqrt 3}{2}\bigg)\bigg)$

$(A)\; \large\frac{1}{2} \\ (B)\; -\large\frac{1}{2} \\ (C)\; 1 \\ (D)\;\large\frac{3}{2} $

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A)
Toolbox:
  • $\cos^{-1} (-x) = \pi - \cos^{-1}x$
  • Principal interval of cos is \([0,\pi]\)
  • $(\cos\large\frac{\pi}{6}=\large\frac{\sqrt{3}}{2}) \rightarrow$ $\cos^{-1} \large (\frac{\sqrt 3}{2}) $$= \large \frac{\pi}{6}$
  • $\cos\; - \large (\frac{\pi}{3}) $$ = \frac{1}{2}$
Given: $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))$
We know that $\cos^{-1} (-x) = \pi - \cos^{-1}x$
$\Rightarrow$ $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))\; $$= \cos \large ( \frac{\pi}{2} -$$ \pi + \cos^{-1} \large (\frac{\sqrt 3}{2}))$ = $\cos \large ( $$\cos^{-1} \large (\frac{\sqrt 3}{2}) - \frac{\pi}{2})$
We know that $\cos^{-1} \large (\frac{\sqrt 3}{2}) $$= \large \frac{\pi}{6}$
$\Rightarrow$ $\cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))\; $ $= \cos \large (\frac{\pi}{6} - \frac{\pi}{2})$$ = \cos - \large (\frac{\pi}{3})$
We know that: $\cos - \large (\frac{\pi}{3}) $$= \frac{1}{2}$
$\Rightarrow \cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))$ $= \large \frac{1}{2}$
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