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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2014 · Mathematics · Trigonometric Ratios & Identities

If A (60^ ^ -A) (60^ ^ +A)=k 3 A , then what is k equal to?

If sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=ksin⁡3A\sin \mathrm{A} \sin \left(60^{{}^{\circ}}-\mathrm{A}\right) \sin \left(60^{{}^{\circ}}+\mathrm{A}\right)=k \sin 3 \mathrm{ A}, then what is kk equal to?

Options

  1. A14\frac{1}{4}
  2. B12\frac{1}{2}
  3. C1
  4. D4

Correct answer

A. 14\frac{1}{4}

Step-by-step solution

sin⁡A⋅(sin⁡260∘−sin⁡2A)=ksin⁡3Asin⁡A((32)2−sin⁡2A)=ksin⁡3Asin⁡A(34−sin⁡2A)=ksin⁡3A3sin⁡A−4sin⁡3A4=ksin⁡3A\begin{aligned} \sin \mathrm{A} \cdot\left(\sin ^2 60^{{}^{\circ}}-\sin ^2 \mathrm{ A}\right) & =\mathrm{k} \sin 3 \mathrm{ A} \\ \sin \mathrm{ A}\left(\left(\frac{\sqrt{3}}{2}\right)^2-\sin ^2 \mathrm{ A}\right) & =k \sin 3 \mathrm{ A} \\ \sin \mathrm{ A}\left(\frac{3}{4}-\sin ^2 \mathrm{ A}\right) & =k \sin 3 \mathrm{ A} \\ \frac{3 \sin \mathrm{ A}-4 \sin ^3 \mathrm{ A}}{4} & =k \sin 3 \mathrm{ A}\end{aligned}sin⁡3A4=ksin⁡3A14k=1k=14\begin{aligned} \frac{\sin 3 \mathrm{ A}}{4} & =\frac{k \sin 3 \mathrm{ A}}{1} \\ 4 k & =1 \\ k & =\frac{1}{4}\end{aligned}$
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