If ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ then $K =$
$\frac{4}{3}$
$\frac{3}{4}$
$\frac{1}{2}$
$2$
If $\frac{{5\pi }}{2} < x < 3\pi $, then the value of the expression $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ is
$cosec^2\theta $ = $\frac{4xy}{(x +y)^2}$ is true if and only if
If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, then $2\beta $ is equal to
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :
The value of $tan^{-1} (\frac{sin2 -1}{cos2})$ is equal to:-