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

In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4(A B)$. If the area of $\triangle CAB$ is $2 \sqrt{3}-3$ unit $^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

The value of ,$\sqrt 3 \, cosec\, 20^o - sec\, 20^o $ is :

If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:

$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ is equal to

$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $