Question

If $cosecA-\sin A=m$ and $\sec A-\cos A=n$, prove that $(m^{2}n{)}^{2/3}+(m{n}^2{)}^{2/3}=1$.

Answer 1

$cosecA-\sin A=m$

$\sec A-\cos A=n$

$\frac{1}{\sin A}-\sin A=m$ and $\frac{1}{\cos A}-\cos A=n$

$\frac{1-si{n}^{2}A}{\sin A}=m$ and $\frac{1-co{s}^{2}A}{\cos A}=n$

$\frac{co{s}^{2}A}{sinA}=m$ and $\frac{si{n}^{2}A}{cosA}=n$

Therefore,

$(m^{2}n{)}^{2/3}+(m{n}^2{)}^{2/3}=(\frac{co{s}^{4}A}{si{n}^{2}A}\times \frac{si{n}^{2}A}{cosA}{)}^{2/3}+(\frac{co{s}^{2}A}{sinA}\times \frac{si{n}^{4}A}{co{s}^{2}A}{)}^{2/3}$

$=(co{s}^{3}A{)}^{2/3}+(si{n}^{3}A{)}^{2/3}$

$=co{s}^{2}A+si{n}^{2}A=1$