Question

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

Answer 1

Given : $cosec\theta -\sin \theta =a^3$ and $\sec \theta -\cos \theta =b^3$

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

$\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }=a^3$

$\Rightarrow a={\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{1}{3}}$

Similarly ; $b={\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{1}{3}}$

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

$\Rightarrow {\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{4}{3}}{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{2}{3}}+{\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{2}{3}}{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{4}{3}}$

$\Rightarrow (cos\theta {)}^{8/3-2/3}.(sin\theta {)}^{4/3-4/3}+(sin\theta {)}^{8/3-2/3}.(cos\theta {)}^{4/3-4/3}$

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow a^2{b}^2(a^2+b^2)=1$