if (cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3,

Solution:

We have $(\operatorname{cosec} \theta-\sin \theta)=a^{3}$

$=>a^{3}=\left(\frac{1}{\sin \theta}-\sin \theta\right)$

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

$\therefore a=\frac{\cos ^{\frac{2}{3}} \theta}{\sin ^{\frac{1}{3}} \theta}$

Again, $(\sec \theta-\cos \theta)=b^{3}$

$=>b^{3}=\left(\frac{1}{\cos \theta}-\cos \theta\right)$

$=\frac{\left(1-\cos ^{2} \theta\right)}{\cos \theta}$

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

$\therefore b=\frac{\sin ^{\frac{2}{3}} \theta}{\cos ^{\frac{1}{3}} \theta}$

Now, LHS $=a^{2} b^{2}\left(a^{2}+b^{2}\right)$

$=a^{4} b^{2}+a^{2} b^{4}$

$=a^{3}\left(a b^{2}\right)+\left(a^{2} b\right) b^{3}$

$=\frac{\cos ^{2} \theta}{\sin \theta} \times\left[\frac{\cos \frac{2}{3} \theta}{\sin \frac{1}{3} \theta} \times \frac{\sin ^{\frac{4}{3}} \theta}{\cos ^{\frac{2}{3}} \theta}\right]+\left[\frac{\cos \frac{4}{3} \theta}{\sin ^{\frac{2}{3}} \theta} \times \frac{\sin \frac{2}{3} \theta}{\cos ^{\frac{1}{3}} \theta}\right] \times \frac{\sin ^{2} \theta}{\cos \theta}$

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

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

$=$ RHS

Hence proved.