Question

If \[x = a{\cos ^3}\theta \], \[y = b{\sin ^3}\theta \] then
A) \[{\left( {\dfrac{a}{x}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{b}{y}} \right)^{\dfrac{2}{3}}} = 1\]
B) \[{\left( {\dfrac{b}{x}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{a}{y}} \right)^{\dfrac{2}{3}}} = 1\]
C) \[{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1\]
D) \[{\left( {\dfrac{x}{b}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{a}} \right)^{\dfrac{2}{3}}} = 1\]

Answer 1

Hint: Here in this question, we have to eliminate the theta ‘\[\theta \]’ in the given trigonometric expressions. For this, first we need to solve the given two expression for \[\cos \theta \] and \[\sin \theta \] then add the two expressions and further simplify by using a standard trigonometric identity i.e., \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] to get the required solution.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function, also called circular function.
Consider the given two expressions:
\[x = a{\cos ^3}\theta \] ----(1)
\[y = b{\sin ^3}\theta \] ----(2)
Here, we need to eliminate the ‘\[\theta \]’ .
Consider the equation (1)
\[ \Rightarrow \,\,\,x = a{\cos ^3}\theta \]
Divide both side by ‘\[a\]’, then we have
\[ \Rightarrow \,\,\,\,\dfrac{x}{a} = {\cos ^3}\theta \]
Taking cube roots on both the sides, then
\[ \Rightarrow \,\,\,\,{\left( {\dfrac{x}{a}} \right)^{\dfrac{1}{3}}} = \cos \theta \]
Taking square on both the sides, then we get
\[ \Rightarrow \,\,\,\,{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} = {\cos ^2}\theta \] ----(3)
Now consider the equation (2)
\[ \Rightarrow \,\,\,y = b{\sin ^3}\theta \]
Divide both side by ‘\[b\]’, then we have
\[ \Rightarrow \,\,\,\,\dfrac{y}{b} = {\sin ^3}\theta \]
Taking cube roots on both sides, then
\[ \Rightarrow \,\,\,\,{\left( {\dfrac{y}{b}} \right)^{\dfrac{1}{3}}} = \sin \theta \]
Taking square on both the sides, then we get
\[ \Rightarrow \,\,\,\,{\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\sin ^2}\theta \] -----(4)
Add both the equation (3) and (4), then
\[ \Rightarrow \,\,\,\,{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = {\cos ^2}\theta + {\sin ^2}\theta \] ------(5)
On applying the standard trigonometric identity, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], then equation (5) becomes.
\[\therefore \,\,\,\,{\left( {\dfrac{x}{a}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{y}{b}} \right)^{\dfrac{2}{3}}} = 1\]
Hence, it’s a required solution.
Therefore, option (C) is the correct option.

Note:
For this type of question, we have to eliminate the \[\theta \] it means that the two or more equations are combined logically into one equation so that it remains valid and theta \[\left( \theta \right)\] does not appear in the solution. Students should know the basic formulas of trigonometric identities, law of indices and basic arithmetic operations.