If tan theta cdot tan left(120^{circ}-thetari | vedclass

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

Question

(A) We know the identity: $	an 	heta \cdot 	an \left(60^{\circ}-	hetaight) \cdot 	an \left(60^{\circ}+	hetaight) = 	an 3	heta$. Given the

Vedclass · Accepted Answer

$\frac{n \pi}{3}+\frac{\pi}{18}, n \in Z$

Vedclass · Answer

(A) We know the identity: $	an 	heta \cdot 	an \left(60^{\circ}-	hetaight) \cdot 	an \left(60^{\circ}+	hetaight) = 	an 3	heta$. Given the expression: $	an 	heta \cdot 	an \left(120^{\circ}-	hetaight) \cdot 	an \left(120^{\circ}+	hetaight) = \frac{1}{\sqrt{3}}$. Using the identity $	an(180^{\circ} - A) = -	an A$,we have $	an(120^{\circ} - 	heta) = -	an(60^{\circ} + 	heta)$ and $	an(120^{\circ} + 	heta) = -	an(60^{\circ} - 	heta)$. Substituting these: $	an 	heta \cdot [-	an(60^{\circ} + 	heta)] \cdot [-	an(60^{\circ} - 	heta)] = 	an 	heta \cdot 	an(60^{\circ} + 	heta) \cdot 	an(60^{\circ} - 	heta) = 	an 3	heta$. Thus,$	an 3	heta = \frac{1}{\sqrt{3}}$. Since $	an 3	heta = 	an \frac{\pi}{6}$,the general solution is $3	heta = n\pi + \frac{\pi}{6}$. Dividing by $3$,we get $	heta = \frac{n\pi}{3} + \frac{\pi}{18}, n \in Z$.

If $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$,then $\theta$ is equal to

Similar Questions

If the orthocentre and circumcentre of a triangle $ABC$ are at equal distances from the side $BC$ and lie on the same side of $BC$,then the value of $\tan B \tan C$ is equal to:

Let $\theta, \phi \in [0, 2\pi]$ be such that $2 \cos \theta(1-\sin \phi) = \sin^2 \theta \left(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}\right) \cos \phi - 1$,$\tan (2\pi - \theta) > 0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

If $3 \sin \theta + 4 \cos \theta = 3$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\sin 2 \theta = $

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

The larger of $\cos (\log \theta)$ and $\log (\cos \theta)$,if $e^{-\pi / 2} < \theta < \pi / 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module