Question

If $\frac{9x}{\cos \theta }+\frac{5y}{\sin \theta }=56$ and $\frac{9x\sin \theta }{{\cos }^2\theta }-\frac{5y\cos \theta }{{\sin }^2\theta }=0$, then the value of $\left[\right(9x{)}^{2/3}+(5y{)}^{2/3}{]}^3$ is

Answer 1

Given, $\frac{9x}{\cos \theta }+\frac{5y}{\sin \theta }=56$ .....(1)

Also given, $\frac{9x\sin \theta }{{\cos }^2\theta }-\frac{5y\cos \theta }{{\sin }^2\theta }=0$

$9x{\sin }^3\theta =5y{\cos }^3\theta$.

$\Rightarrow \frac{{\cos }^3\theta }{9x}=\frac{{\sin }^3\theta }{5y}=k^3$ (say)

$\Rightarrow \cos \theta =k(9x{)}^{1/3}$ and $sin\theta =k(5y{)}^{1/3}$ ....(2)

Squaring and adding, we get

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

$\Rightarrow k^2\left[\right(9x{)}^{2/3}+\left(5y{)}^{2/3}\right]=1$

$\Rightarrow k^2=\frac{1}{\left[\right(9x{)}^{2/3}+\left(5y{)}^{2/3}\right]}$

Substitute the value from (2) in (1), we get

$\frac{9x}{k(9x{)}^{1/3}}+\frac{5y}{k(5y{)}^{1/3}}=56$ (by (1))

$\Rightarrow (9x{)}^{2/3}+(5y{)}^{2/3}=56k$

$\Rightarrow \left[\right(9x{)}^{2/3}+(5y{)}^{2/3}{]}^2=(56{)}^2{k}^2$

$\Rightarrow \left[\right(9x{)}^{2/3}+(5y{)}^{2/3}{]}^2=\frac{(56{)}^2}{(9x{)}^{2/3}+(5y{)}^{2/3}}$

$\Rightarrow \left[\right(9x{)}^{2/3}+(5y{)}^{2/3}{]}^3=(56{)}^2=3136$.