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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

What is ( 9^ ^ - 9^ ^ ) equal to?

What is \left(\sin 9^{{}^{\circ}}-\cos 9^{{}^{\circ}}\right) equal to?

Options

  1. A−5−52-\frac{\sqrt{5-\sqrt{5}}}{2}
  2. B−5−32-\frac{\sqrt{5-\sqrt{3}}}{2}
  3. C5−52\frac{\sqrt{5-\sqrt{5}}}{2}
  4. D5−54\frac{\sqrt{5-\sqrt{5}}}{4}

Correct answer

C. 5−52\frac{\sqrt{5-\sqrt{5}}}{2}

Step-by-step solution

sin⁡9∘−cos⁡9∘=2(sin⁡9∘−cos⁡9∘)2=2(12sin⁡9∘−12cos⁡9∘)=2[sin⁡(45∘−9∘)]=2sin⁡36∘=2[2(5−5)4]=25−54=5−52.\begin{aligned} & \sin 9^{{}^{\circ}}-\cos 9^{{}^{\circ}}=\sqrt{2} \frac{\left(\sin 9^{{}^{\circ}}-\cos 9^{{}^{\circ}}\right)}{\sqrt{2}} \\ & =\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin 9^{{}^{\circ}}-\frac{1}{\sqrt{2}} \cos 9^{{}^{\circ}}\right) \\ & =\sqrt{2}\left[\sin \left(45^{{}^{\circ}}-9^{{}^{\circ}}\right)\right] \\ & =\sqrt{2} \sin 36^{{}^{\circ}} \\ & =\sqrt{2}\left[\sqrt{2} \frac{(\sqrt{5-\sqrt{5}})}{4}\right] \\ & =2 \frac{\sqrt{5-\sqrt{5}}}{4} \\ & =\frac{\sqrt{5-\sqrt{5}}}{2}. \end{aligned}
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