Question

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

• A. $\cot 3x$
• B. $\tan 3x$
• C. $3\tan 3x$
• D. $\frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}$

Answer 1

Answer: (D)

$\frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}$

$\to \cot x+\frac{\cot {60}^{0}cotx-1}{\cot {60}^0+\cot x}+\frac{\cot {120}^0\cot x-1}{\cot {120}^0+cotx}$

$\to cotx+\frac{\frac{cotx}{\sqrt{3}}-1}{\frac{1}{\sqrt{3}}+cotx}+\frac{\frac{-1}{\sqrt{3}}\cot x-1}{\frac{-1}{\sqrt{3}}+\cot x}$

$\to \cot x+\frac{8\cot x}{1-3{\cot }^{2}x}=\frac{3{\cot }^{3}x-9cotx}{3{\cot }^{2}x-1}$

$\to \frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}$