(d) Given that $\tan \theta – \cot \theta = a$…..$(i)$

and $\sin \theta + \cos \theta = b$…..$(ii)$

Now ${({b^2} – 1)^2}({a^2} + 4)$

$ = {\left\{ {{{(\sin \theta + \cos \theta )}^2} – 1} \right\}^2}\left\{ {{{(\tan \theta – \cot \theta )}^2} + 4} \right\}$

$ = {[1 + \sin 2\theta – 1]^2}[{\tan ^2}\theta + {\cot ^2}\theta – 2 + 4]$

$ = {\sin ^2}2\theta ({\rm{cose}}{{\rm{c}}^2}\theta + {\sec ^2}\theta )$

$ = 4{\sin ^2}\theta {\cos ^2}\theta \left[ {\frac{1}{{{{\sin }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right] = 4$. 

Trick : Obviously the value of expression ${({b^2} – 1)^2}({a^2} + 4)$ is independent of $\theta $,

therefore put any suitable value of $\theta $.

Let $\theta = 45^\circ $, we get $a = 0,\;b = \sqrt 2 $

so that ${[{(\sqrt 2 )^2} – 1]^2}$ $({0^2} + 4) = 4.$