જો $\theta $ એ લઘુકોણ છે અને $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, તો $\tan \theta = . . .$
${x^2} - 1$
$\sqrt {{x^2} - 1} $
$\sqrt {{x^2} + 1} $
${x^2} + 1$
જો $tan\ 80^o = a$ અને $tan47^o = b$ હોય તો $tan37^o$ =
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
જો $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta ) = . . . .$
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ થાય તો