If x = 8(sin θ + cos θ) and y = 9(sin θ - cos θ), then the value of \({x^2 \over 8^2} + {y^2 \over 9^2}\) is:

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 1)

Answer 1

Given:

x = 8(sin θ + cos θ)

y = 9(sin θ - cos θ)

Formula used:

sin² θ + cos² θ = 1

(a + b)² = a² + b² +2ab

Calculations:

⇒ x = 8(sin θ + cos θ)

⇒ x² = 8²(sin θ + cos θ)²

⇒ x² = 8²(sin2 θ + cos2 θ + 2sinθcosθ)

⇒ x2 = 82(1 + 2sinθcosθ)

similarly,

⇒ y = 9(sin θ - cos θ)

⇒ y2 = 92(sin θ - cos θ)2

⇒ y2 = 92(sin2 θ + cos2 θ - 2sinθcosθ)

⇒ y2 = 92(1 - 2sinθcosθ)

According to the question,

⇒ \({x^2 \over 8^2} + {y^2 \over 9^2}\)

⇒ \({8^2(1 + 2sinθcosθ) \over 8^2} + {9^2(1 - 2sinθcosθ) \over 9^2}\)

⇒ 1 + 2sinθcosθ + 1 - 2sinθcosθ

⇒ 2

⇒ Hence, The value of above equation is 2

Shortcut Trick

If we put, θ = 0° we get x = 8 and y = 9, so

\({x^2 \over 8^2} + {y^2 \over 9^2}\)

⇒ 8²/8² + 9²/9²

⇒ 1 + 1

⇒ 2