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If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b,$ then ${({b^2} - 1)^2}({a^2} + 4)$ is equal to
$2$
$-4$
$± 4$
$4$
Solution
(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 ${[{(\sqrt 2 )^2} – 1]^2}$ $({0^2} + 4) = 4.$