If sqrt{3} tan 2theta + sqrt{3} tan 3theta + | vedclass

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

Question

(B) Given equation: $\sqrt{3} 	an 2	heta + \sqrt{3} 	an 3	heta + 	an 2	heta 	an 3	heta = 1$Rearranging the terms: $\sqrt{3}(	an 2	heta + 	a

Vedclass · Accepted Answer

$\left(n + \frac{1}{6}\right)\frac{\pi}{5}$

Vedclass · Answer

(B) Given equation: $\sqrt{3} 	an 2	heta + \sqrt{3} 	an 3	heta + 	an 2	heta 	an 3	heta = 1$Rearranging the terms: $\sqrt{3}(	an 2	heta + 	an 3	heta) = 1 - 	an 2	heta 	an 3	heta$Dividing both sides by $\sqrt{3}(1 - 	an 2	heta 	an 3	heta)$,we get:$\frac{	an 2	heta + 	an 3	heta}{1 - 	an 2	heta 	an 3	heta} = \frac{1}{\sqrt{3}}$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$	an(2	heta + 3	heta) = 	an\left(\frac{\pi}{6}ight)$$	an 5	heta = 	an\left(\frac{\pi}{6}ight)$The general solution for $	an x = 	an \alpha$ is $x = n\pi + \alpha$.Therefore,$5	heta = n\pi + \frac{\pi}{6}$$	heta = \frac{n\pi}{5} + \frac{\pi}{30} = \left(n + \frac{1}{6}ight)\frac{\pi}{5}$.

If $\sqrt{3} \tan 2\theta + \sqrt{3} \tan 3\theta + \tan 2\theta \tan 3\theta = 1$,then the general value of $\theta$ is

Similar Questions

The number of principal solutions of $\tan 2\theta = 1$ is

The general solution of the equation $\cot \theta - \tan \theta = 2$ is

Values of $\theta$ $(0^\circ < \theta < 360^\circ)$ satisfying $\text{cosec } \theta + 2 = 0$ are

The number of elements in the set $\{x \in [0, 180^{\circ}] : \tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan x \tan(x-50^{\circ})\}$ is . . . . . . .

If $\sin^2 \theta - 2\cos \theta + \frac{1}{4} = 0$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module