Question

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

Answer 1

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