tan, 20^{circ} + tan, 40^{circ} + sqrt{3}, ta | vedclass

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

Question

(C) We know that $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) We know that $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$\sqrt{3} = \frac{	an 40^{\circ} + 	an 20^{\circ}}{1 - 	an 40^{\circ} 	an 20^{\circ}}$.Cross-multiplying gives:$\sqrt{3}(1 - 	an 40^{\circ} 	an 20^{\circ}) = 	an 40^{\circ} + 	an 20^{\circ}$.$\sqrt{3} - \sqrt{3} 	an 40^{\circ} 	an 20^{\circ} = 	an 40^{\circ} + 	an 20^{\circ}$.Rearranging the terms:$	an 20^{\circ} + 	an 40^{\circ} + \sqrt{3} 	an 20^{\circ} 	an 40^{\circ} = \sqrt{3}$.

$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ is equal to

Similar Questions

If $\sin x + \sin y = \frac{1}{2}$ and $\cos x + \cos y = 1$,then $\tan(x + y) = $

If $\theta$ and $\phi$ are angles in the $1^{st}$ quadrant such that $\tan \theta = 1/7$ and $\sin \phi = 1/\sqrt{10}$. Then:

Prove that $\sin^{2} 6x - \sin^{2} 4x = \sin 2x \sin 10x$.

If $\cos \alpha + \cos \beta = \frac{1}{3}$ and $\sin \alpha + \sin \beta = \frac{1}{4}$,then $\cos (\alpha + \beta) = $

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$,where $0 \le \alpha, \beta \le \frac{\pi}{4}$,then $\tan 2\alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module