1+cos 10^{circ}+cos 20^{circ}+cos 30^{circ}= | vedclass

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

Question

(B) हमारे पास है,$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ} = (1+\cos 10^{\circ}) + (\cos 20^{\circ}+\cos 30^{\circ})$ $= 2\cos^2 5^{\circ} +

Vedclass · Accepted Answer

$4 \cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$

Vedclass · Answer

(B) हमारे पास है,$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ} = (1+\cos 10^{\circ}) + (\cos 20^{\circ}+\cos 30^{\circ})$ $= 2\cos^2 5^{\circ} + 2\cos 25^{\circ} \cos 5^{\circ}$ $= 2\cos 5^{\circ} (\cos 5^{\circ} + \cos 25^{\circ})$ $= 2\cos 5^{\circ} (2\cos \frac{25^{\circ}+5^{\circ}}{2} \cos \frac{25^{\circ}-5^{\circ}}{2})$ $= 2\cos 5^{\circ} (2\cos 15^{\circ} \cos 10^{\circ})$ $= 4\cos 5^{\circ} \cos 10^{\circ} \cos 15^{\circ}$

$1+\cos 10^{\circ}+\cos 20^{\circ}+\cos 30^{\circ}=$

Similar Questions

सिद्ध कीजिए कि $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$.

यदि $\cos (A - B) = \frac{3}{5}$ और $\tan A \tan B = 2$ है,तो

$\cos 105^\circ + \sin 105^\circ$ का मान है

सिद्ध कीजिए कि $\frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x$.

यदि $A=35^{\circ}, B=15^{\circ}$ और $C=40^{\circ}$ है,तो $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module