Question
Download Solution PDFयदि x = 8(sin θ + cos θ) और y = 9(sin θ - cos θ) है, तो \({x^2 \over 8^2} + {y^2 \over 9^2}\) का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया:
x = 8(sin θ + cos θ)
y = 9(sin θ - cos θ)
प्रयुक्त सूत्र:
sin2 θ + cos2 θ = 1
(a + b)2 = a2 + b2 +2ab
गणना:
⇒ x = 8(sin θ + cos θ)
⇒ x2 = 82(sin θ + cos θ)2
⇒ x2 = 82(sin2 θ + cos2 θ + 2sinθcosθ)
⇒ x2 = 82(1 + 2sinθcosθ)
उसी प्रकार,
⇒ y = 9(sin θ - cos θ)
⇒ y2 = 92(sin θ - cos θ)2
⇒ y2 = 92(sin2 θ + cos2 θ - 2sinθcosθ)
⇒ y2 = 92(1 - 2sinθcosθ)
प्रश्न के अनुसार,
⇒ \({x^2 \over 8^2} + {y^2 \over 9^2}\)
⇒ \({8^2(1 + 2sinθcosθ) \over 8^2} + {9^2(1 - 2sinθcosθ) \over 9^2}\)
⇒ 1 + 2sinθcosθ + 1 - 2sinθcosθ
⇒ 2
⇒ इसलिए, उपरोक्त समीकरण का मान 2 है।
Last updated on Jul 15, 2025
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