If tan x + tan left( frac{pi }{3} + x right) | vedclass

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

Question

(C) We use the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Given expression: $	an x + 	an \left( \frac{\pi }{3} + x ight) + \

Vedclass · Accepted Answer

$	an 3x = 1$

Vedclass · Answer

(C) We use the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Given expression: $	an x + 	an \left( \frac{\pi }{3} + x ight) + 	an \left( \frac{2\pi }{3} + x ight) = 3$Expanding the terms:$= 	an x + \frac{	an x + \sqrt{3}}{1 - \sqrt{3} 	an x} + \frac{	an x - \sqrt{3}}{1 + \sqrt{3} 	an x}$$= 	an x + \frac{(	an x + \sqrt{3})(1 + \sqrt{3} 	an x) + (	an x - \sqrt{3})(1 - \sqrt{3} 	an x)}{(1 - \sqrt{3} 	an x)(1 + \sqrt{3} 	an x)}$$= 	an x + \frac{(	an x + \sqrt{3} 	an^2 x + \sqrt{3} + 3 	an x) + (	an x - \sqrt{3} 	an^2 x - \sqrt{3} + 3 	an x)}{1 - 3 	an^2 x}$$= 	an x + \frac{8 	an x}{1 - 3 	an^2 x} = \frac{	an x - 3 	an^3 x + 8 	an x}{1 - 3 	an^2 x} = \frac{9 	an x - 3 	an^3 x}{1 - 3 	an^2 x}$$= 3 \left( \frac{3 	an x - 	an^3 x}{1 - 3 	an^2 x} ight) = 3 	an 3x$Given $3 	an 3x = 3$,therefore $	an 3x = 1$.

If $\tan x + \tan \left( \frac{\pi }{3} + x \right) + \tan \left( \frac{2\pi }{3} + x \right) = 3,$ then

Similar Questions

The value of $\cos 105^\circ + \sin 105^\circ$ is:

If $A + B + C = \pi$ and $\cos A = \cos B \cos C$,then $\tan B \tan C$ is equal to

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

If $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$,then the value of the expression $\frac{\sin(A + B)}{\sin 2C}$ is:

The maximum value of $f(x) = \sin x + \cos x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module