$(\sec 2A + 1){\sec ^2}A = $
$\sec A$
$2\sec A$
$\sec 2A$
$2\sec 2A$
Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to
$2\cos x - \cos 3x - \cos 5x = $
If $2\sec 2\alpha = \tan \beta + \cot \beta ,$ then one of the values of $\alpha + \beta $ is
If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $