(d) : $P = \left\{ (a,b):\sec^{2}a - \tan^{2}b = 1 \right\}$ For reflexive : $\sec^{2}a - \tan^{2}a = 1($ true $\forall a)$ For Symmetric : $\sec^{2}b - \tan^{2}a = 1$ L.H.S $1 + \tan^{2}b - \left( \sec^{2}a - 1 \right) = 1 + \tan^{2}b - \sec^{2}a + 1 = - \left( \sec^{2}a - \tan^{2}b \right) + 2 = - 1 + 2 = 1$ So, relation is symmetric. For transitive : If $\sec^{2}a - \tan^{2}b = 1$ and $\sec^{2}b - \tan^{2}c = 1$ $\sec^{2}a - \tan^{2}c = \left( 1 + \tan^{2}b \right) - \left( \sec^{2}b - 1 \right)$\\ $= - 1 + 2 = 1$ Hence, relation $P$ is an equivalence relation.