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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Answers & Comments
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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