\(\frac{1}{p - q}\)
\(\frac{1}{q - 1}\)
\(\frac{1}{q + 1}\)
1 + 0
\(\frac{1}{1 - q}\)
Correct answer is C
Pq + 1 = q2......(i)
t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii)
p = \(\frac{q^2 - 1}{q}\)
Sub for p in equation (ii)
t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\)
t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\)
t = \(\frac{q - 1}{q^2 - 1}\)
= \(\frac{q - 1}{(q + 1)(q - 1)}\)
= \(\frac{1}{q + 1}\)