The value of $\cos 15^\circ - \sin 15^\circ $ is equal to
$\frac{1}{{\sqrt 2 }}$
$\frac{1}{2}$
$ - \frac{1}{{\sqrt 2 }}$
$0$
If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-
If $A$ and $B$ are complimentary angles, then :
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
The value of ${\cos ^2}\,{10^o}\,\, - \,\cos \,\,{10^o}\,\cos \,\,{50^o}\, + \,{\cos ^2}\,{50^o}$ is