If x=a(Sinθ+Cosθ) and y=b(Sin θ-Cosθ), then the value of (x²/a²)+(y²/b²) is:

  1. 3
  2. 4
  3. 2
  4. 1
Anurag Mishra Professor Asked on 10th December 2015 in Maths.
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1 Answer(s)

Answer:  (3) 2

Explanation:-
X = a(Sinθ + Cosθ)
Square on both side,
X² = a² (Sin²θ + Cos²θ + 2 Sinθ Cosθ)
X²= a² (1 + 2 Sinθ Cosθ)

Y = b (Sinθ – Cosθ)
Square on both side,
Y² = b² (Sin²θ + Cos²θ – 2 Sinθ Cosθ)
Y²  = b² (1 – 2 Sinθ Cosθ)

Now, X²/a² + Y²/b²
Put the value of Y²  = b² (1 – 2 Sinθ Cosθ) & X²= a² (1 + 2 Sinθ Cosθ)

Then, a² (1 + 2 Sinθ Cosθ)/a² + b² (1 – 2 Sinθ Cosθ)/b²
1 + 2 Sinθ Cosθ + 1 – 2 Sinθ Cosθ
1 + 1
= 2

Hence, the correct answer is option (3) 2.

Anurag Mishra Professor Answered on 10th December 2015.
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