• question_answer $\tan \left( \frac{A}{2} \right)$is equal to A)  $\frac{\tan A}{(1+secA)}$B)  $\frac{1}{(cosecA+cotA)}$C) $\frac{\tan A}{(1+cosecA)}$D) $\frac{1}{(secA+\cot A)}$

$\tan \frac{A}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}$ $=\frac{2\sin \frac{A}{2}.\cos \frac{A}{2}}{2{{\cos }^{2}}\frac{A}{2}}$ [Multiplying numerator and denominator by $2\cos \frac{A}{2}$] $=\frac{\sin A}{1+\cos A}$ $=\frac{\frac{\sin A}{\cos A}}{\frac{1+\cos A}{\cos A}}=\frac{\tan A}{\sec A+1}$ or When $A={{60}^{o}},$ $\tan \frac{A}{2}=\tan {{30}^{o}}=\frac{1}{\sqrt{3}}$ and $\frac{\tan A}{1+\sec A}=\frac{\tan {{60}^{o}}}{1+\sec {{60}^{o}}}$ $=\frac{\sqrt{3}}{1+2}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$