Question

If $\csc \theta -\sin \theta =a^3$, $\sec \theta -\cos \theta =b^3$ P.T $a^2{b}^2(a^2+b^2)=1$

Answer 1

$\csc \theta -\sin \theta =a^3$

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

$\Rightarrow \sec \theta -\cos \theta =b^3$

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

$a^4{b}^2+a^2{b}^4=\frac{{\cos }^{\frac{8}{3}}\theta }{{\sin }^{\frac{4}{3}}\theta }\cdot \frac{{\sin }^{\frac{4}{3}}\theta }{{\cos }^{\frac{2}{3}}\theta }+\frac{{\cos }^{\frac{4}{3}}\theta }{{\sin }^{\frac{2}{3}}\theta }\cdot \frac{{\sin }^{\frac{8}{3}}\theta }{{\cos }^{\frac{4}{3}}\theta }$

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

$=1$

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