Question

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

Answer 1

$cosecA-\sin A=a^3$

$\frac{1}{\sin A}-\sin A=a^3$

$\frac{1-{\sin }^{2}A}{\sin A}=a^3=\frac{{\cos }^{2}A}{\sin A}\Rightarrow a={\left[\frac{{\cos }^{2}A}{\sin A}\right]}^{\frac{1}{3}}$

$\sec A-\cos A=b^3$

$\frac{1-{\cos }^{2}A}{\cos A}=b^3\Rightarrow b^3=\frac{{\sin }^{2}A}{\cos A}\Rightarrow b={\left[\frac{{\sin }^{2}A}{\cos A}\right]}^{\frac{1}{3}}$

$a^2{b}^2[a^2+b^2]=a^3{b}^3[\frac{a}{b}+\frac{b}{a}]$

$\frac{a}{b}=\frac{(\cos A{)}^{\frac{2}{3}}}{(\sin A{)}^{\frac{1}{3}}}\times \frac{(\cos A{)}^{\frac{1}{3}}}{(\sin A{)}^{\frac{2}{3}}}=\frac{\cos A}{\sin A}$

$=\frac{{\cos }^{2}A}{\sin A}\times \frac{{\sin }^{2}A}{\cos A}\times \frac{{\sin }^{2}A}{\cos A}[\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}]$

$=\cos A\sin A\left[\frac{1}{\cos A\sin A}\right]=1$

Hence proved.