$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
- A
$\cos A$
- B
$0$
- C
$\sqrt 3 \sin A$
- D
$\sqrt 3 \cos A$
Similar Questions
$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
- [IIT 1977]
If $A + B + C = \pi ,$ then $\cos \,\,2A + \cos \,\,2B + \cos \,\,2C = $
If $A + B + C = \pi ,$ then $\cos \,\,2A + \cos \,\,2B + \cos \,\,2C = $
In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is
In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is