If $ a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta=m, a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta=n $, prove that $ (m+n)^{2 / 3}+(m-n)^{2 / 3}=2 a^{2 / 3} $

Given:

 \( a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta=m, a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta=n \)

To do:

We have to prove that \( (m+n)^{2 / 3}+(m-n)^{2 / 3}=2 a^{2 / 3} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta=m$.........(i)

$a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta=n$.........(ii)

Adding (i) and (ii), we get,

$m+n=a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta+a \sin ^{3} \theta+3 a \cos ^{2} \theta \sin \theta$

$=a[\cos ^{3} \theta+3 \cos ^{2} \theta \sin \theta+3 \cos \theta \sin ^{2} \theta+\sin ^{3} \theta]$ $=a(\cos \theta+\sin \theta)^{3}$        (Since $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$)

Subtracting (ii) from (i), we get,

$m-n=a \cos ^{3} \theta+3 a \cos \theta \sin ^{2} \theta-a\sin ^{3} \theta-3 a \cos ^{2} \theta \sin \theta$

$=a[\cos ^{3} \theta-3 \cos ^{2} \theta \sin \theta+3 \cos \theta \sin ^{2} \theta-\sin ^{3} \theta]$

$=a(\cos \theta-\sin \theta)^{3}$    [Since $(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$]

This implies,

$(m+n)^{\frac{2}{3}}+(m-n)^{\frac{2}{3}}=[a(\cos \theta+\sin \theta)^{3}]^{\frac{2}{3}}+[a(\cos \theta-\sin \theta)^{3}]^{\frac{2}{3}}$

$=a^{\frac{2}{3}}(\cos \theta+\sin \theta)^{2}+a^{\frac{2}{3}}(\cos \theta-\sin \theta)^{2}$

$=a^{\frac{2}{3}}[(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}]$

$=a^{\frac{2}{3}}[(\cos ^{2} \theta+\sin ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta]$

$=a^{\frac{2}{3}}[2(\cos ^{2} \theta+\sin ^{2} \theta)]$

$=a^{\frac{2}{3}}(2 \times 1)$

$=2 a^{\frac{2}{3}}$

Hence proved.