Question

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

Answer 1

$$\begin{gathered} \mathrm{cosec}x-\sin x=a^3 \\ \therefore \frac{1}{\sin x}-\sin =a^3 \\ \Rightarrow \frac{1-{\sin }^{2}x}{\sin x}=a^3 \\ \Rightarrow \frac{{\cos }^{2}x}{\sin x}=a^3 \\ \Rightarrow a={\left(\frac{{\cos }^{2}x}{\sin x}\right)}^{\frac{1}{3}}....\left(\mathrm{i}\right) \\ \mathrm{Also},\sec x-\cos x=b^3 \\ \Rightarrow \frac{1}{\cos x}-\cos =b^3 \\ \Rightarrow \frac{1-{\cos }^{2}x}{\cos x}=b^3 \\ \Rightarrow \frac{{\sin }^{2}x}{\cos x}=b^3 \\ \Rightarrow b={\left(\frac{{\sin }^{2}x}{\cos x}\right)}^{\frac{1}{3}}.....\left(\mathrm{ii}\right) \\ \mathrm{Now},\mathrm{LHS}=a^2{b}^2\left(a^2+b^2\right)={\left(ab\right)}^2\left(a^2+b^2\right) \\ ={\left[{\left(\frac{{\cos }^{2}x}{\sin x}\right)}^{\frac{1}{3}}{\left(\frac{{\sin }^{2}x}{\cos x}\right)}^{\frac{1}{3}}\right]}^2\left[{\left({\left(\frac{{\cos }^{2}x}{\sin x}\right)}^{\frac{1}{3}}\right)}^2+{\left({\left(\frac{{\sin }^{2}x}{\cos x}\right)}^{\frac{1}{3}}\right)}^2\right] \\ ={\left(\sin x\cos x\right)}^{\frac{2}{3}}\left[\frac{{\left({\cos }^{2}x\right)}^{{\displaystyle \frac{2}{3}}}}{{\left(\sin x\right)}^{{\displaystyle \frac{2}{3}}}}+\frac{{\left({\sin }^{2}x\right)}^{{\displaystyle \frac{2}{3}}}}{{\left(\cos x\right)}^{{\displaystyle \frac{2}{3}}}}\right] \\ ={\left(\sin x\cos x\right)}^{\frac{2}{3}}\left[\frac{{\left({\cos }^{3}x\right)}^{\frac{2}{3}}+{\left({\sin }^{3}x\right)}^{\frac{2}{3}}}{{\left(\sin x\right)}^{\frac{2}{3}}{\left(\cos x\right)}^{\frac{2}{3}}}\right] \\ ={\left(\sin x\cos x\right)}^{\frac{2}{3}}\left[\frac{{\cos }^{2}x+{\sin }^{2}x}{{\left(\sin x\cos x\right)}^{\frac{2}{3}}}\right] \\ =1=\mathrm{RHS} \end{gathered}$$