tan^{-1} left[ frac{sqrt{2}}{sqrt{3}} cos lef | vedclass

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(C) We know that $\sin^{-1}(1/\sqrt{2}) = \pi/4$.Substituting this into the expression,we get $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left(

Vedclass · Accepted Answer

$-\pi/6$

Vedclass · Answer

(C) We know that $\sin^{-1}(1/\sqrt{2}) = \pi/4$.Substituting this into the expression,we get $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \cdot \frac{\pi}{4} ight) ight]$.Since $\cos(5\pi/4) = \cos(\pi + \pi/4) = -\cos(\pi/4) = -1/\sqrt{2}$.Substituting this value,the expression becomes $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cdot \left( -\frac{1}{\sqrt{2}} ight) ight]$.This simplifies to $	an^{-1} \left( -\frac{1}{\sqrt{3}} ight)$.Since $	an^{-1}(-x) = -	an^{-1}(x)$ and $	an^{-1}(1/\sqrt{3}) = \pi/6$,the final result is $-\pi/6$.Thus,option $C$ is correct.

$\tan^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \sin^{-1} \frac{1}{\sqrt{2}} \right) \right] = \dots$

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