The value of cot, x + cot, (60^circ + x) + co | vedclass

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(A) Let $f(x) = \cot x + \cot(60^\circ + x) + \cot(120^\circ + x)$.Using the identity $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we have:$\co

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$3\, \cot\, 3x$

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(A) Let $f(x) = \cot x + \cot(60^\circ + x) + \cot(120^\circ + x)$.Using the identity $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we have:$\cot(60^\circ + x) + \cot(120^\circ + x) = \frac{\sin(180^\circ + 2x)}{\sin(60^\circ + x)\sin(120^\circ + x)} = \frac{-\sin 2x}{\sin(60^\circ + x)\sin(60^\circ - x)}$.Using $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$,the denominator is $\sin^2 60^\circ - \sin^2 x = \frac{3}{4} - \sin^2 x = \frac{3 - 4\sin^2 x}{4}$.So,$\cot(60^\circ + x) + \cot(120^\circ + x) = \frac{-2\sin 2x}{\frac{3 - 4\sin^2 x}{2}} = \frac{-4\sin x \cos x}{\frac{3 - 4\sin^2 x}{2}} = \frac{-8\sin x \cos x}{3 - 4\sin^2 x}$.Now,$f(x) = \frac{\cos x}{\sin x} - \frac{8\sin x \cos x}{3 - 4\sin^2 x} = \frac{\cos x(3 - 4\sin^2 x) - 8\sin^2 x \cos x}{\sin x(3 - 4\sin^2 x)}$.$f(x) = \frac{3\cos x - 4\sin^2 x \cos x - 8\sin^2 x \cos x}{3\sin x - 4\sin^3 x} = \frac{3\cos x - 12\sin^2 x \cos x}{3\sin x - 4\sin^3 x}$.Dividing numerator and denominator by $\cos^3 x$ or using $\cot 3x = \frac{\cot^3 x - 3\cot x}{3\cot^2 x - 1}$,we find $f(x) = 3\cot 3x$.

The value of $\cot\, x + \cot\, (60^\circ + x) + \cot\, (120^\circ + x)$ is equal to:

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