Question

If cosecA - sinA=m and secA-cosA=n,prove that (mn2)2/3 + (m2n)2/3=1

Answer 1

$(m^{2}n)^{\dfrac{2}{3}}+(mn^{2} )^{\dfrac{2}{3}}=\cos ^{2} A+\sin^{2} A=1,$ proved.

Step-by-step explanation:

We have,

$\csc A-\sin A=m$ and $\sec A-\cos A=n$

To prove that,  $(mn^2)^{\dfrac{2}{3}}  (m^2n)^{\dfrac{2}{3}} =1$

∴ $m^{2} n=(\csc A-\sin A)^{2} .(\sec A-\cos A)$

$=\dfrac{(1-\sin^{2} A)^{2}}{\sin^{2} A}.\dfrac{(1-\cos^{2} A)^{2}}{\cos A}$

$=\dfrac{\cos^{4} A}{\sin^{2} A} \times \dfrac{\sin^{2} A}{\cos A}<strong>=\cos^{3} A$

∴$(m^{2}n)^{\dfrac{1}{3}} =\cos A$     ...(1)

$mn^{2} =(\csc A-\sin A)^{2} .((\sec A-\cos A))^{2}$

$=\dfrac{(1-\sin^{2} A)}{\sin A}.\dfrac{(1-\cos^{2} A)^{2}}{\cos^{2} A}$

$=\dfrac{\cos^{2} A}{\sin A} \times \dfrac{\sin^{4} A}{\cos^{2} A}=\sin^{3} A$

∴$(mn^{2} )^{\dfrac{1}{3}} =\sin A$   .....(2)

Squaring and adding (1) and (2), we get

$(m^{2}n)^{\dfrac{2}{3}}+(mn^{2} )^{\dfrac{2}{3}}=\cos ^{2} A+\sin^{2} A=1$, proved.