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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2023 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Let β be the GM of α and α ; γ be the AM of α and α...

Paragraph: Let sin⁡β\sin \beta be the GM of sin⁡α\sin \alpha and cos⁡α;tan⁡γ\cos \alpha ; \tan \gamma be the AM of sin⁡α\sin\alpha and cos⁡α\cos \alpha
Question: What is the value of of sec⁡2γ\sec 2 \gamma ?

Options

  1. A3−sin⁡2α5+2sin⁡2α\frac{3-\sin 2 \alpha}{5+2 \sin 2 \alpha}
  2. B5+sin⁡2α3−sin⁡2α\frac{5+\sin 2 \alpha}{3-\sin 2 \alpha}
  3. C3−2sin⁡2α4+sin⁡2α\frac{3-2 \sin 2 \alpha}{4+\sin 2 \alpha}
  4. D3−sin⁡2α4+3sin⁡2α\frac{3-\sin 2 \alpha}{4+3 \sin 2 \alpha}

Correct answer

B. 5+sin⁡2α3−sin⁡2α\frac{5+\sin 2 \alpha}{3-\sin 2 \alpha}

Step-by-step solution

sec⁡2γ=1cos⁡2γ=1+tan⁡2γ1−tan⁡2γ\sec 2 \gamma=\frac{1}{\cos 2 \gamma}=\frac{1+\tan ^2 \gamma}{1-\tan ^2 \gamma} Hence Sec⁡2ystan⁡γ=Sin⁡α+Cos⁡α2=1+(sin⁡α+cos⁡α2)21−(sin⁡α+cos⁡α2)2=4+(sin⁡α+cos⁡α)24−(sin⁡α+cos⁡α)2=4+(1+sin⁡2α)4−(1+sin⁡2α)=5+sin⁡2α3−sin⁡2α\operatorname{Sec}_2 y \operatorname{stan} \gamma=\frac{\operatorname{Sin} \alpha+\operatorname{Cos} \alpha}{2}=\frac{1+\left(\frac{\sin \alpha+\cos \alpha}{2}\right)^2}{1-\left(\frac{\sin \alpha+\cos \alpha}{2}\right)^2}=\frac{4+(\sin \alpha+\cos \alpha)^2}{4-(\sin \alpha+\cos \alpha)^2}=\frac{4+(1+\sin 2 \alpha)}{4-(1+\sin 2 \alpha)}=\frac{5+\sin 2 \alpha}{3-\sin 2 \alpha}.
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