Question

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

Answer 1

$a^3=\frac{{\cos }^{2}x}{sinx}$ and $b^3=\frac{{\sin }^{2}x}{cosx}$

$\to a^4{b}^2+a^2{b}^4=a^2{b}^2(a^2+b^2)$

$\to a^4{b}^2+a^2{b}^4={(\frac{{\cos }^{8}x}{{\sin }^{4}x}.\frac{{\sin }^{4}x}{{\cos }^{2}x})}^{1/3}+{(\frac{{\sin }^{8}x}{{\cos }^{2}x}.\frac{{\cos }^{4}x}{{\sin }^{2}x})}^{1/6}$

$\to {\cos }^{2}x+{\sin }^{2}x=1=a^2{b}^2(a^2+b^2)$