If tan theta + tan left(theta + frac{pi}{3}ri | vedclass

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

Question

(B) Given the equation: $	an 	heta + 	an \left(	heta + \frac{\pi}{3}ight) + 	an \left(	heta + \frac{2\pi}{3}ight) = 3$. Using the identity $

Vedclass · Accepted Answer

$	an 3	heta$

Vedclass · Answer

(B) Given the equation: $	an 	heta + 	an \left(	heta + \frac{\pi}{3}ight) + 	an \left(	heta + \frac{2\pi}{3}ight) = 3$. Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we expand the terms: $	an 	heta + \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} + \frac{	an 	heta - \sqrt{3}}{1 + \sqrt{3} 	an 	heta} = 3$. Combining the fractions: $	an 	heta + \frac{(	an 	heta + \sqrt{3})(1 + \sqrt{3} 	an 	heta) + (	an 	heta - \sqrt{3})(1 - \sqrt{3} 	an 	heta)}{1 - 3 	an^2 	heta} = 3$. Simplifying the numerator: $	an 	heta + \frac{	an 	heta + \sqrt{3} 	an^2 	heta + \sqrt{3} + 3 	an 	heta + 	an 	heta - \sqrt{3} 	an^2 	heta - \sqrt{3} + 3 	an 	heta}{1 - 3 	an^2 	heta} = 3$. $	an 	heta + \frac{8 	an 	heta}{1 - 3 	an^2 	heta} = 3$. $\frac{	an 	heta - 3 	an^3 	heta + 8 	an 	heta}{1 - 3 	an^2 	heta} = 3$. $\frac{9 	an 	heta - 3 	an^3 	heta}{1 - 3 	an^2 	heta} = 3$. $3 \left( \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta} ight) = 3$. Since $	an 3	heta = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$,we get $3 	an 3	heta = 3$,which implies $	an 3	heta = 1$. Thus,option $B$ is correct.

If $\tan \theta + \tan \left(\theta + \frac{\pi}{3}\right) + \tan \left(\theta + \frac{2\pi}{3}\right) = 3$,then which of the following is equal to $1$?

Similar Questions

$\frac{\sec 8\theta - 1}{\sec 4\theta - 1}$ is equal to

$2\cos x - \cos 3x - \cos 5x = $

If $2 \cos \theta = x + \frac{1}{x}$,then $2 \cos 3 \theta = $

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module