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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2017 · Mathematics · Trigonometric Ratios & Identities

If A=(3)/(5) , where 450^ ^ < A < 540^ ^ , then A 2 is equal to

If sin⁡A=35\sin \mathrm{A}=\frac{3}{5}, where 450∘<A<540∘450^{{}^{\circ}} < \mathrm{A} < 540^{{}^{\circ}}, then cos⁡A2\cos \frac{\mathrm{A}}{2} is equal to

Options

  1. A110\frac{1}{\sqrt{10}}
  2. B−310-\sqrt{\frac{3}{10}}
  3. C310\frac{\sqrt{3}}{\sqrt{10}}
  4. DNone of the above

Correct answer

D. None of the above

Step-by-step solution

Sin⁡A=35;450∘<A<540∘⇒225∘<A2<270∘cos⁡A=−45(∵A lies in  Q2)\begin{aligned} & \operatorname{Sin} \mathrm{A}=\frac{3}{5} ; 450^{{}^{\circ}} < \mathrm{A} < 540^{{}^{\circ}} \\ & \Rightarrow 225^{{}^{\circ}} < \frac{\mathrm{A}}{2} < 270^{{}^{\circ}} \\ & \cos \mathrm{A}=\frac{-4}{5} \quad(\because \mathrm{ A} \text{ lies\text{ in } } \mathrm{Q} 2)\end{aligned}∴cos⁡2A2=1+cos⁡A2=110⇒cos⁡A2=−110(∵A2 lies in Q3 )\begin{aligned} & \therefore \cos ^2 \frac{\mathrm{ A}}{2}=\frac{1+\cos \mathrm{A}}{2}=\frac{1}{10} \\ & \Rightarrow \cos \frac{\mathrm{ A}}{2}=\frac{-1}{\sqrt{10}}\left(\because \frac{\mathrm{ A}}{2} \text{ lies in Q3 }\right)\end{aligned}$
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