Question

If $\csc \theta -\sin \theta =a^3$ and $\sec \theta -\cos \theta =b^3$, prove that $a^2{b}^2(a^2+b^2)=1$

Answer 1

$a^3=\frac{{\cos }^2\theta }{\sin \theta },$ $b^3=\frac{{\sin }^2\theta }{\cos \theta }$

$a^2{b}^2(a^2+b^2)={\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{2}{3}}{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{2}{3}}[{\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{2}{3}}+{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{2}{3}}]$

$a^2{b}^2(a^2+b^2)={\left(\sin \theta \cos \theta \right)}^{\frac{2}{3}}\times \frac{({\cos }^2\theta +{\sin }^2\theta )}{{\left(\cos \theta \sin \theta \right)}^{\frac{2}{3}}}$

$a^2{b}^2(a^2+b^2)=1$ $\left(Proved\right)$