Question

if x is equal to a cos cube theta and Y is equal to b sin cube theta prove that X by a to the power 2 by 3 + 5 by b to the power 3 is equal to 1

Answer 1

by shortcut method

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by shortcut method plzz like and rateif u like method

Answer 2

Answer:

 $(\frac{x}{a} )^{\frac{2}{3} }$ + $(\frac{y}{b} )^{\frac{2}{3} }$ = cos² + sin² = 1 (the proof is given in the explanation)

Step-by-step explanation:

• Given information: x = a cos³ and y = b sin³
• We have to prove,  $(\frac{x}{a} )^{\frac{2}{3} }$ + $(\frac{y}{b} )^{\frac{2}{3} }$ = 1
• First taking the term = $(\frac{x}{a} )^{\frac{2}{3} }$

                                          = $(\frac{acos^{3}\theta}{a} )^{\frac{2}{3} }$   [ as x = a cos³ ]

                                          = $({cos\theta} )^{\frac{2}{3} *3 }$

                                          = $({cos\theta} )^{2}$ = cos²

• Now the 2nd term = $(\frac{y}{b} )^{\frac{2}{3} }$

                                        = $(\frac{bsin^{3}\theta}{b} )^{\frac{2}{3} }$   [ as y = b sin³]

                                        = $({sin\theta} )^{\frac{2}{3} *3 }$

                                         = $({sin\theta} )^{2}$ = sin²

• By adding  $(\frac{x}{a} )^{\frac{2}{3} }$ and  $(\frac{y}{b} )^{\frac{2}{3} }$ we get,  $(\frac{x}{a} )^{\frac{2}{3} }$ + $(\frac{y}{b} )^{\frac{2}{3} }$ = cos² + sin²

                                                                                    = 1 (proved)

 [ Applying trigonometry, for a right angled triangle sin = $\frac{height}{hypotenuse}$

                                                                           and    cos = $\frac{base}{hypotenuse}$

 Now, sin² + cos²

    = $(\frac{height}{hypotenuse})^{2}$ + $(\frac{base}{hypotenuse})^{2}$

 = $\frac{hypotenuse^{2} }{hypotenuse^{2} }$ [applying Pythagoras theorem, base²+height²=hypotenuse²]

 = 1 ]