The value of $\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)$ is equal to
$-\frac{1}{2}$
$-1$
$-\frac{1}{3}$
$-\frac{1}{4}$
Prove that $\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x$
The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :
If $A, B, C$ are angles of a triangle, then $\sin 2A + \sin 2B - \sin 2C$ is equal to
The value of $cot\, x + cot\, (60^o + x) + cot\, (120^o + x)$ is equal to :
The value of the expression $(sinx + cosecx)^2 + (cosx + secx)^2 - ( tanx + cotx)^2$ wherever defined is equal to