7・3 If $x$ is a real number satisfying $\sec x-\operatorname{tg} x=2$, then the value of $\sec x+\operatorname{tg} x$ is
(A)0.1.
(B) 0.2.
(C) 0.3.
(D) 0.4.
(E)0.5.
(50th American High School Mathematics Examination, 1999)
[Solution 1] Given $\frac{1-\sin x}{\cos x}=2$, multiply both sides by $1+\sin x$ to get
$$
\frac{1-\sin ^{2} x}{\cos x}=2(1+\sin x)
$$

which simplifies to $\frac{1+\sin x}{\cos x}=\frac{1}{2}=0.5$, noting that $1-\sin ^{2} x=\cos ^{2} x$. Therefore, the answer is $(E)$.
[Solution 2] From $\sec ^{2} x=1+\operatorname{tg}^{2} x$,
we have $\sec ^{2} x-\operatorname{tg}^{2} x=1$,
thus $(\sec x-\operatorname{tg} x)(\sec x+\operatorname{tg} x)=1$.
Given $\sec x-\operatorname{tg} x=2$, it follows that $\sec x+\operatorname{tg} x=\frac{1}{2}$. Therefore, the answer is $(E)$.