Question

$\huge{\underline{\underline{\bf Question}}}$

if $\sf x = acos³\theta$ and $\sf y = bsin³\theta$ then show that :-
$\sf \bigg( \dfrac{x}{a} \bigg)^{ \dfrac{2}{3} } + \bigg( \dfrac{y}{b} \bigg)^{ \dfrac{2}{3} } = 1$
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Answer 1

Answer:

We say that copper is the positive pole and zinc is the negative one, but in reality, the transition of electrons will happen against electrostatic forces, not following them: the positive electrode, copper, will become negatively charged from the extra electrons, at the expense of the negative electrode, zinc whic

Answer 2

Given :-

x = acos³θ and y = bsin³θ

To Find :-

Prove that ( x/a )^2/3 + ( y/b )^2/3 = 1

Used Concepts :-

• Sin² x + Cos² x = 1
• ( x^m )^n = x^m × n

Solution :-

Let us start with the Given i.e

x = a cos³θ

=> x/a = cos³θ ----- ( i )

y = b sin³θ

=> y/b = sin³θ ----- ( ii )

=> Now we will prove which we have to do :)

=> ( x/a )^2/3 + ( y/b )^2/3 = 1

Let LHS i.e ( x/a )^2/3 + ( y/b )^2/3

=> ( Cos³θ )^2/3 + ( Sin³θ )^2/3

{ Using ( i ) and ( ii ) }

=> { ( Cosθ )³ }^2/3 + { ( Sinθ )³ }^2/3

=> ( Cosθ ) ² + ( Sinθ ) ²

{ Because , ( x^m )^n = x^m × n }

=> Cos²θ + Sin²θ

{ Because Cos² x + Sin² x = 1 }

=> 1

LHS = RHS

Henceforth , Proved !!