If \[\tan \theta -\cot \theta =a\] and \[\sin \theta +\cos \theta =b,\] then \[{{({{b}^{2}}-1)}^{2}}({{a}^{2}}+4)\] is equal to [WB JEE 1979]

Given that \[\TAN \theta -\COT \theta =a\] ?..(i) and \[\sin \theta +\cos \theta =b\] ?..(ii) Now \[{{({{b}^{2}}-1)}^{2}}({{a}^{2}}+4)\] \[={{\left\{ {{(\sin \theta +\cos \theta )}^{2}}-1 \right\}}^{2}}\left\{ {{(\tan \theta -\cot \theta )}^{2}}+4 \right\}\] \[={{[1+\sin 2\theta -1]}^{2}}[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta -2+4]\] \[={{\sin }^{2}}2\theta (\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta )\] \[=4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left[ \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } \right]=4\]. 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 \[{{[{{(\sqrt{2})}^{2}}-1]}^{2}}\] \[({{0}^{2}}+4)=4.\]