Question

If $\csc \theta -\sin \theta =a^{3}4\sec \theta -\cos \theta =b^3$
Prove that $a^2{b}^2(a^2+b^2)=1$

Answer 1

$co\sec \theta -\sin \theta =a^3$

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

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

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

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

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

Now

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

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

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

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

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

$\frac{{\sin }^{4/3}\theta }{{\cos }^{2/3}\theta }=b^2...\left(5\right)$

from (4) and (5)

$a^2\times b^2=\frac{{\cos }^{4/3}\theta }{{\sin }^{2/3}\theta }\times \frac{{\sin }^{4/3}\theta }{{\cos }^{2/3}\theta }$

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

from (4) and (5)

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

Now

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