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

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(C) 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 ha

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$3\, cot\, 3x$

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(C) 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:$cot\, (60^\circ + x) + cot\, (120^\circ + x) = \frac{\sin(180^\circ + 2x)}{\sin(60^\circ + x)\sin(120^\circ + x)}$.Since $\sin(180^\circ + 2x) = -\sin 2x$ and $\sin(60^\circ + x)\sin(120^\circ + x) = \sin^2 60^\circ - \sin^2 x = \frac{3}{4} - \sin^2 x$,$f(x) = cot\, x - \frac{\sin 2x}{\frac{3}{4} - \sin^2 x} = \frac{\cos x}{\sin x} - \frac{2\sin x \cos x}{\frac{3-4\sin^2 x}{4}} = \frac{\cos x}{\sin x} - \frac{8\sin x \cos x}{3-4\sin^2 x}$.$f(x) = \frac{\cos x(3-4\sin^2 x) - 8\sin^2 x \cos x}{\sin x(3-4\sin^2 x)} = \frac{3\cos x - 4\sin^2 x \cos x - 8\sin^2 x \cos x}{3\sin x - 4\sin^3 x} = \frac{3\cos x - 12\sin^2 x \cos x}{3\sin x - 4\sin^3 x}$.Dividing numerator and denominator by $\cos^3 x$ or using the triple angle formula:$f(x) = \frac{3\cos x(1 - 4\sin^2 x)}{3\sin x - 4\sin^3 x} = \frac{3\cos x(1 - 4(1-\cos^2 x))}{\sin 3x} = \frac{3\cos x(4\cos^2 x - 3)}{\sin 3x} = \frac{3(4\cos^3 x - 3\cos x)}{\sin 3x} = \frac{3\cos 3x}{\sin 3x} = 3\, cot\, 3x$.

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

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