If m = a cos³β and n = b sin³β, then find the value of $$\left( \frac{m}{a} \right)^{\frac{2}{3}} + \left( \frac{n}{b} \right)^{\frac{2}{3}}$$.

Correct option is C

Given: ​

m = a cos³β

n = b sin³β 

Formula Used: 

sin²β  + cos²β = 1 

Solution: ​

​$$m = a \cos³β \implies \cosβ = \left(\frac ma \right)^{\frac{1}{3}} \\ \ \\ n = b \sin³β \implies \sinβ = \left(\frac nb \right)^{\frac{1}{3}}$$ 

Now, 

​$$\sin^2β + \cos^2β = 1 \\ \ \\ \left[\left(\frac ma \right)^{\frac13} \right]^2 +\left[\left(\frac nb \right)^{\frac13} \right]^2 = 1 \\ \ \\ \left(\frac ma \right)^{\frac23} +\left(\frac nb \right)^{\frac23} = 1$$​​