Question

If (cosec θ − sin θ) = a³ and (sec θ − cos θ) = b³, prove that $a^2{b}^2(a^2+b^2)=1.$

Answer 1

$$\begin{gathered} \text{We have}\left(\cos \text{ec}\theta \text{-sin}\theta \right)\text{=}{a}^3 \\ \begin{array}{l}\text{=>}{a}^3=(\frac{1}{\sin \theta }-\sin \theta )\\ \end{array} \\ \begin{array}{l}\text{=>}{a}^3=\frac{(1-{\sin }^2\theta )}{\sin \theta }\\ \end{array}=\frac{{\cos }^2\theta }{\sin \theta } \\ \therefore a=\frac{{\cos }^{\frac{2}{3}}\theta }{{\sin }^{\frac{1}{3}}\theta } \\ \begin{array}{l}\text{Again,}(\sec \theta -\cos \theta )=b^3\\ \end{array} \\ =>b^3=(\frac{1}{\cos \theta }-\cos \theta ) \\ =\frac{(1-{\cos }^2\theta )}{\cos \theta } \\ =\frac{{\sin }^2\theta }{\cos \theta } \\ \therefore b=\frac{{\sin }^{\frac{2}{3}}\theta }{{\cos }^{\frac{1}{3}}\theta } \\ \begin{array}{l}\text{Now, LHS=}{a}^2{b}^2(a^2+b^2)\\ \end{array} \\ \text{=}{a}^4{b}^2+a^2{b}^4 \\ =a^3\left(a{b}^2\right)+\left(a^{2}b\right)b^3 \\ \begin{array}{ll}=\frac{{\cos }^2\theta }{\sin \theta }\times \left[\frac{{\cos }^{\frac{2}{3}}\theta }{{\sin }^{\frac{1}{3}}\theta }\times \frac{{\sin }^{\frac{4}{3}}\theta }{{\cos }^{\frac{2}{3}}\theta }\right]+\left[\frac{{\cos }^{\frac{4}{3}}\theta }{{\sin }^{\frac{2}{3}}\theta }\times \frac{{\sin }^{\frac{2}{3}}\theta }{{\cos }^{\frac{1}{3}}\theta }\right]\times \frac{{\sin }^2\theta }{\cos \theta }& \\ =\frac{{\cos }^2\theta }{\sin \theta }\times \sin \theta +\cos \theta \times \frac{{\sin }^2\theta }{\cos \theta }& \end{array} \\ {\text{=cos}}^2\theta +{\sin }^2\theta =1 \\ =\text{RHS} \\ \text{Hence proved}\text{.} \end{gathered}$$