The value of $\sum_{r-1}^{18} cos^2(5r)^o,$ where $x^o $ denotes the $x$ degree, is equals to
If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for
$\tan 75^\circ - \cot 75^\circ = $
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 $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to