$7 \cdot 6 \operatorname{ctg} 10+\operatorname{tg} 5$ 的值为  
(A) csc5.  
(B) $\csc 10$.  
(C) sec5.  
(D) $\sec 10$.  
(E) $\sin 15$.  
(40th American High School Mathematics Examination, 1989)

The value of $7 \cdot 6 \operatorname{ctg} 10+\operatorname{tg} 5$ is  
(A) csc5.  
(B) $\csc 10$.  
(C) sec5.  
(D) $\sec 10$.  
(E) $\sin 15$.  
(40th American High School Mathematics Examination, 1989)
\begin{array}{l}\text { [Solution] From } \operatorname{ctg} 2 \alpha+\operatorname{tg} \alpha=\frac{1}{\operatorname{tg} 2 \alpha}+\operatorname{tg} \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{2 \operatorname{tg} \alpha}+\operatorname{tg} \alpha \\ =\frac{1+\operatorname{tg}^{2} \alpha}{2 \operatorname{tg} \alpha}=\frac{1}{\sin 2 \alpha}=\csc 2 \alpha, \\ \therefore \quad \operatorname{ctg} 10+\operatorname{tg} 5=\csc 10 . \\ \end{array}
Therefore, the correct choice is $(B)$.