Question

Mates please answer this question......

No spams please........ ​

Answer 1

Step-by-step explanation:

Squaring both the equations and then adding ,

x²/a² cos²θ + y²/b² sin²θ + 2xy/ab sinθ.cosθ + x²/a² sin²θ + y²/b² cos²θ - 2xy/ab sinθ.cosθ = 2

⇒x²/a² (cos²θ + sin²θ) + y²/b² (sin²θ + cos²θ ) = 2

⇒x²/a² × 1 + y²/b² × 1 = 2 [ ∵ sin²x + cos²x = 1 from trigonometric identities ]

∴ x²/a² + y²/b² = 2 , hence proved

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• x/a (cos θ) + y/b (sin θ) = 1 -----(1)
• x/a (sin θ) - y/b (cos θ) = 1 -----(2)

$\Large{\underline{\mathfrak{\bf{Prove\:That}}}}$

• x²/a² + y²/b² = 2

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

Squaring both side of equ(1)

➠ [x/a (cos θ) + y/b (sin θ)]² = 1²

➠ (x/a)² cos² θ + (y/b)² sin² θ + 2*(x/a)*(y/b) sin θ * cos θ = 1 -------------(3)

Again, Squaring both side of equ(2)

➠[ x/a (sin θ) - y/b (cos θ)]² = 1²

➠ (x/a)² sin² θ + (y/b)² cos² θ - 2*(x/a)*(y/b)* cos θ * sin θ = 1 --------------(4)

Add equ(3) & equ(4)

➠ [(x/a)² cos² θ + (y/b)² sin² θ + 2*(x/a)*(y/b) sin θ * cos θ ] + [ (x/a)² sin² θ + (y/b)² cos² θ - 2*(x/a)*(y/b)* cos θ * sin θ] = 1 + 1

➠ (x/a)²( cos² θ + sin² θ) + (y/b)² (cos² θ + sin² θ) = 2

[ ★ ( cos² θ + sin² θ) = 1 ]

➠ (x/a)² * 1 + (y/b)² * 1 = 2

➠ (x²/a²) + (y²/b²) = 2

That's Proved.

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