Let f(x) = cot left( sin^{-1} sqrt{frac{2}{3 | vedclass

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(C) Given $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$.We know that $\cos 2x = 2\cos^2 x - 1$,so $3 + \cos 2x = 3 + 2\cos^2 x -

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$\frac{\sqrt{3}}{2}$

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(C) Given $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} ight)$.We know that $\cos 2x = 2\cos^2 x - 1$,so $3 + \cos 2x = 3 + 2\cos^2 x - 1 = 2 + 2\cos^2 x = 2(1 + \cos^2 x)$.Substituting this into the expression,we get $\sqrt{\frac{2}{2(1 + \cos^2 x)}} = \frac{1}{\sqrt{1 + \cos^2 x}}$.However,let us simplify the argument of $\cot$ differently. Let $	heta = \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}}$. Then $\sin 	heta = \sqrt{\frac{2}{3 + \cos 2x}}$.$\sin^2 	heta = \frac{2}{3 + \cos 2x} \implies \csc^2 	heta = \frac{3 + \cos 2x}{2} = \frac{3 + 2\cos^2 x - 1}{2} = 1 + \cos^2 x$.Since $\cot^2 	heta = \csc^2 	heta - 1$,we have $\cot^2 	heta = (1 + \cos^2 x) - 1 = \cos^2 x$.Thus,$f(x) = \cot 	heta = \sqrt{\cos^2 x} = |\cos x|$.At $x = \frac{2\pi}{3}$,$\cos x = -\frac{1}{2}$,so $f(x) = |\cos x| = -\cos x$ in the neighborhood of $\frac{2\pi}{3}$.$f'(x) = \frac{d}{dx}(-\cos x) = \sin x$.$f'\left( \frac{2\pi}{3} ight) = \sin \left( \frac{2\pi}{3} ight) = \frac{\sqrt{3}}{2}$.

Let $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$. Then,the value of $f'\left( \frac{2\pi}{3} \right)$ is:

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