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

  • A

    $\frac{1}{2}$

  • B

    $1\over3$

  • C

    $\frac{1}{4}$

  • D

    $\frac{1}{8}$

Similar Questions

The value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is $............$.

  • [JEE MAIN 2023]

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

${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $

$\sin 4\theta $ can be written as