Answer
(a). $ a^{2} - 1$
(b). $\tan^{2}\theta$
Work Step by Step
(a). Given expression is-
$ (a+1) (a-1)$
= $ a^{2} -a + a -1$
( By FOIL method)
= $ a^{2} - 1$
ALTERNATE METHOD-
Recall that $(A+B) (A-B) = A^{2} - B^{2}$
Therefore $ (a+1) (a-1)$ = $ a^{2} - 1^{2}$ = $ a^{2} - 1$
(b). Given expression is-
$ (\sec\theta+1) (\sec\theta-1)$
= $ \sec^{2}\theta -\sec\theta + \sec\theta -1$
( By FOIL method)
= $ \sec^{2}\theta - 1$
= $\tan^{2}\theta$
( Recall second Pythagorean identity)
ALTERNATE METHOD-
Recall that $(A+B) (A-B) = A^{2} - B^{2}$
Therefore $ (\sec\theta+1) (\sec\theta-1)$ = $ \sec^{2}\theta - 1^{2}$
= $ \sec^{2}\theta - 1$
= $\tan^{2}\theta$
