If $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$then ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta )$ is equal to

The value of $tan^{-1} (\frac{sin2 -1}{cos2})$ is equal to:-

If $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ find $\tan 2A$ in terms of $\tan B$ and show that

Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $

$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi  \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi  \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :

If $cosA + cosB = cosC,\ sinA + sinB = sinC$ then the value of expression $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ is