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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

54^ ^ can be expressed as

tan⁡54∘\tan 54^{{}^{\circ}} can be expressed as

Options

  1. Asin⁡9∘+cos⁡9∘sin⁡9∘−cos⁡9∘\frac{\sin 9^{{}^{\circ}}+\cos 9^{{}^{\circ}}}{\sin 9^{{}^{\circ}}-\cos 9^{{}^{\circ}}}
  2. Bsin⁡9∘−cos⁡9∘sin⁡9∘+cos⁡9∘\frac{\sin 9^{{}^{\circ}}-\cos 9^{{}^{\circ}}}{\sin 9^{{}^{\circ}}+\cos 9^{{}^{\circ}}}
  3. Ccos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘\frac{\cos 9^{{}^{\circ}}+\sin 9^{{}^{\circ}}}{\cos 9^{{}^{\circ}}-\sin 9^{{}^{\circ}}}
  4. Dsin⁡36∘cos⁡36∘\frac{\sin 36^{{}^{\circ}}}{\cos 36^{{}^{\circ}}}

Correct answer

C. cos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘\frac{\cos 9^{{}^{\circ}}+\sin 9^{{}^{\circ}}}{\cos 9^{{}^{\circ}}-\sin 9^{{}^{\circ}}}

Step-by-step solution

tan⁡54∘=tan⁡(45∘+9∘)=tan⁡45∘+tan⁡9∘1−tan⁡45∘⋅tan⁡9∘=1+tan⁡9∘1−tan⁡9∘=1+sin⁡9∘cos⁡9∘1−sin⁡9∘cos⁡9∘=cos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘\begin{aligned} & \tan 54^{{}^{\circ}}=\tan \left(45^{{}^{\circ}}+9^{{}^{\circ}}\right) \\ & =\frac{\tan 45^{{}^{\circ}}+\tan 9^{{}^{\circ}}}{1-\tan 45^{{}^{\circ}} \cdot \tan 9^{{}^{\circ}}}=\frac{1+\tan 9^{{}^{\circ}}}{1-\tan 9^{{}^{\circ}}} \\ & =\frac{1+\frac{\sin 9^{{}^{\circ}}}{\cos 9^{{}^{\circ}}}}{1-\frac{\sin 9^{{}^{\circ}}}{\cos 9^{{}^{\circ}}}}=\frac{\cos 9^{{}^{\circ}}+\sin 9^{{}^{\circ}}}{\cos 9^{{}^{\circ}}-\sin 9^{{}^{\circ}}}\end{aligned}
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