Question

If $\cos ec\theta -\sin \theta =a^3,\sec \theta -\cos \theta =b^3$
then $a^2{b}^2(a^2+b^2)$ is equal to

Answer 1

Given that,

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

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

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

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

$\cot \theta \cos \theta =a^3$ $\tan \theta \sin \theta =b^3$

$a^2=(\cos \theta \cot \theta {)}^{2/3}$ $b^2=(\tan \theta \sin \theta {)}^{2/3}$

$a^2{b}^2=\left(\sin \theta {)}^{2/3}\right(\cos \theta {)}^{2/3}$

$a^2+b^2=\frac{(\cos \theta {)}^{4/3}}{(\sin \theta {)}^{2/3}}+\frac{(\sin \theta {)}^{4/3}}{(\cos \theta {)}^{2/3}}=\frac{{\sin }^2\theta +{\cos }^2\theta }{\left(\sin \theta {)}^{2/3}\right(\cos \theta {)}^{2/3}}=\frac{1}{(\sin \theta \cos \theta {)}^{2/3}}$

$a^2{b}^2(a^2+b^2)=(\sin \theta \cos \theta {)}^{2/3}\times \frac{1}{(\sin \theta \cos \theta {)}^{2/3}}=1$

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