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

Consider $cosec\theta -\sin \theta =a^3$

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

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

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

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

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

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

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

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

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

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

$\frac{{\sin }^{\frac{4}{2}}\theta }{{\cos }^{\frac{2}{2}}\theta }=b^2$ ……$\left(4\right)$

Multiply $\left(2\right)$ & $\left(4\right)$ we get

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

${\sin }^{\frac{2}{2}}\theta {\cos }^{\frac{2}{2}}\theta =a^2{b}^2$ ……$\left(5\right)$

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

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

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

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