If \(p q+1=q^{2}\) and \(t=\frac{1}{p}-\frac{1}{p q}\), expresstin terms of \(q\) ____________
A. \(\frac{1}{1-p q}\)
B. \(\frac{1}{q}-1\)
C. \(\frac{1}{q}+1\)
D. \(1+q\)
E. \(\frac{1}{1-q}\)
Correct Answer: C
Explanation
\(p q+1=q^{2}\) \(t=\frac{1}{p}-\frac{1}{p q}\)rom (1). \(\frac{q^{2}-1}{q}=p\)rom \((2) \cdot t=\frac{1}{p}-\frac{1}{p q}\)
\(I=\frac{1}{p}\left(1-\frac{1}{q}\right) \cdot p=\frac{1}{1}\left(1-\frac{1}{q}\right) \cdots\)
we equate (3) and (4) and eliminate \(t\)
$$
\begin{aligned}
\frac{q^{2}-1}{q} &=\frac{1}{1}\left(1-\frac{1}{q}\right) \\
1 &=\frac{q}{\left(q^{2}-1\right)}\left(1-\frac{1}{q}\right): \\
1 &=\frac{q-1}{q^{2}-1}=\frac{q-1}{(q-1)(q+1)} \\
&=\frac{1}{q+1}
\end{aligned}
$$