If $\theta = 3\, \alpha$ and $sin\, \theta =$ $\frac{a}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$. The value of the expression , $a \,cosec\, \alpha - b \,sec\, \alpha$ is
$\frac{1}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$
$2 \sqrt {{a^2}\,\, + \,\,{b^2}}$
$a + b$
none
$2\cos x - \cos 3x - \cos 5x = $
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :
$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $
The value of $\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ is
The value of $\frac{{3 + \cot \,7\,{6^ \circ }\,\cot \,{{16}^ \circ }}}{{\cot \,{{76}^ \circ } + \cot \,{{16}^ \circ }}}$ is :