Question

If $cosec\theta -\sin \theta =a^3$ and $\sec \theta -\cos \theta =b^3$, then $a^2{b}^2(a^2+b^2)=$

Answer 1

$\csc \theta -\sin \theta =a^3$
$\frac{1}{\sin \theta }-\sin \theta =a^3$
$\frac{1-{\sin }^2\theta }{\sin \theta }=a^3$
$a^3=\frac{{\cos }^2\theta }{\sin \theta }=\cot \theta \cos \theta$ ....... $\left(i\right)$
Also, $\sec \theta -\cos \theta =b^3$
$\frac{1}{\cos \theta }-\cos \theta =b^3$
$b^3=\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$
$b^3=\tan \theta \sin \theta$ ....... $\left(ii\right)$

Consider,
$a^2{b}^2\left(a^2{+b}^2\right)={\left[\cot \theta \cos \theta \tan \theta \sin \theta \right]}^{2/3}\left[a^2{+b}^2\right]$
$={\left[\sin \theta \cos \theta \right]}^{{}^{2/3}}[{\left(\cot \theta \cos \theta \right)}^{2/3}+{\left(\tan \theta \sin \theta \right)}^{2/3}]$
$={\left[\sin \theta \cos \theta \right]}^{{}^{2/3}}[{\left[\frac{{\cos }^2\theta }{\sin \theta }\right]}^{2/3}+{\left[\frac{{\sin }^2\theta }{\cos \theta }\right]}^{2/3}]$
${=\left[\sin \theta \cos \theta \right]}^{{}^{2/3}}\left[\frac{{\cos }^{\frac{4}{3}}\theta {\cos }^{\frac{2}{3}}\theta +{\sin }^{\frac{4}{3}}\theta {\sin }^{\frac{2}{3}}\theta }{{\left[\sin \theta \cos \theta \right]}^{{}^{2/3}}}\right]$
$={\cos }^2\theta +{\sin }^2\theta$
$=1$