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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$3\, \cot\, 3x$

Vedclass · Answer

(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:

Similar Questions

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to:

If $\sec \theta = 1\frac{1}{4}$,then $\tan \frac{\theta }{2} = $

$\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta} = $

If $a \sin^2 x + b \cos^2 x = c$,$b \sin^2 y + a \cos^2 y = d$ and $a \tan x = b \tan y$,then $\frac{a^2}{b^2}$ is equal to

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module