Question

If $a{\cos }^3\theta +3a\cos \theta {\sin }^2\theta =m$ and $a{\sin }^3\theta +3a{\cos }^2\theta \sin \theta =n$ then ${(m+n)}^{\frac{2}{3}}+{(m-n)}^{\frac{2}{3}}$ is equal to

• A. $a^{\frac{2}{3}}$
• B. $a^{\frac{4}{3}}$
• C. $2a^{\frac{2}{3}}$
• D. $2a^{\frac{4}{3}}$

Answer 1

Answer: (C)

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

Given : $a{\cos }^3\theta +3a\cos \theta {\sin }^2\theta =m$.....(1)

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

Adding equation (1) and (2) we get,

$\therefore m+n=a({\cos }^3\theta +{\sin }^3\theta )+3a\cos \theta \sin \theta (\cos \theta +\sin \theta )=a{(\cos \theta +\sin \theta )}^3$
Similarly, we get
$m-n=a{(\cos \theta -\sin \theta )}^3$

$={(m+n)}^{\frac{2}{3}}+{(m-n)}^{\frac{2}{3}}$

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

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

After solving the above two brackets we get,

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

Hence option ${}^{\prime }{C}^{\prime }$ is the answer.