Question
Download Solution PDFजर \(\sqrt{2}\)sec2 θ - 4sec θ + 2\(\sqrt{2}\) = 0 असेल, तर sin2 θ + tan2 θ चे मूल्य काय आहे?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFगणना:
√2 sec2 - 4sec θ + 2\(√{2}\) = 0
⇒ √2 sec2 - 2sec θ - 2sec θ + 2\(√{2}\) = 0
⇒ √2sec θ(sec θ - √2) - 2(sec θ - √2) = 0
⇒ (sec θ - √2)(√2sec θ- 2) = 0
⇒ sec θ = √2
⇒ θ = 45°
⇒ sin2 θ + tan2 θ = sin2 45° + tan2 45°
⇒ 1/2 + 1 = 3/2
∴ पर्याय 4 हे योग्य उत्तर आहे.
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