• # question_answer If ${{\theta }_{1}},\,\,{{\theta }_{2}}$ are the solutions of the equation$2{{\tan }^{2}}\theta -4\tan \theta +1=0$, then $\tan ({{\theta }_{1}}+{{\theta }_{2}})$is equal to A) $-4$                           B) $4$C) $1$                             D) $2$

Let $\tan {{\theta }_{1}},\,\,\tan {{\theta }_{2}}$ be the roots of the$2{{\tan }^{2}}\theta -4\tan \theta +1=0$ equation. Thus $\tan {{\theta }_{1}}+\tan {{\theta }_{2}}=4/2=2;\,\,\tan {{\theta }_{1}}\tan {{\theta }_{2}}=1/2$. Now$\tan ({{\theta }_{1}}+{{\theta }_{2}})=[(\tan {{\theta }_{1}}+\tan {{\theta }_{2}})/(1-\tan {{\theta }_{1}}\tan {{\theta }_{2}})]$$=2/[1-(1/2)]=4$.