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

In the graph of the function $\sqrt{3} \sin x + \cos x$,the maximum distance of a point from the $x$-axis is:

$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = $

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

If $\tan \theta + \cot \theta = 2$,then $\sin \theta =$

$\frac{\cos A}{1 - \sin A} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module