The correct option is B Statement - 1 is true, Statement - 2 is false.
The truth table for the logical statements, involved in statement 1, is as follows:
pq∼qp↔∼q∼(p↔∼q)p↔qTTFFTTTFTTFFFTFTFFFFTFTT
We observe the columns for ∼(p↔∼q) and p↔q are identical, therefore
∼(p↔∼q) is equivalent to p↔q
But ∼(p↔∼q) is not a tautology as all entries in its column are not T.
∴ Statement - 1 is true but statement - 2 is false.