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) આપણે જાણીએ છીએ કે $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.$	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ સૂત્રનો ઉ

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.$	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ સૂત્રનો ઉપયોગ કરતા:$\sqrt{3} = \frac{	an 40^{\circ} + 	an 20^{\circ}}{1 - 	an 40^{\circ} 	an 20^{\circ}}$.ગુણાકાર કરતા:$\sqrt{3}(1 - 	an 40^{\circ} 	an 20^{\circ}) = 	an 40^{\circ} + 	an 20^{\circ}$.$\sqrt{3} - \sqrt{3} 	an 40^{\circ} 	an 20^{\circ} = 	an 40^{\circ} + 	an 20^{\circ}$.પદોને ગોઠવતા:$	an 20^{\circ} + 	an 40^{\circ} + \sqrt{3} 	an 20^{\circ} 	an 40^{\circ} = \sqrt{3}$.

$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ ની કિંમત શું થાય?

Similar Questions

ધારો કે $\cos (\alpha+\beta)=\frac{4}{5}$ અને $\sin (\alpha-\beta)=\frac{5}{13}$,જ્યાં $0 \leq \alpha, \beta \leq \frac{\pi}{4}$,તો $\tan 2 \alpha=$

ધારો કે $\tan \alpha = \frac{a}{a+1}$ અને $\tan \beta = \frac{1}{2a+1}$, તો $\alpha + \beta$ શું થાય?

$\frac{\tan 52^{\circ} - \tan 38^{\circ}}{\tan 14^{\circ}} = $

જો $A+B=\frac{\pi}{4}$ હોય,તો $\frac{\cos B-\sin B}{\cos B+\sin B}=$

$\cos^2 \left( \frac{\pi}{6} + \theta \right) - \sin^2 \left( \frac{\pi}{6} - \theta \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module