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

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

What is 17^ ^ - 17^ ^ 17^ ^ + 17^ ^ equal to?

What is \frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}$ equal to?

Options

  1. Atan⁡34∘\tan 34^{{}^{\circ}}
  2. Bcot⁡34∘\cot 34^{{}^{\circ}}
  3. Ctan⁡62∘\tan 62^{{}^{\circ}}
  4. Dcot⁡62∘\cot 62^{{}^{\circ}}

Correct answer

D. cot⁡62∘\cot 62^{{}^{\circ}}

Step-by-step solution

 Let I=cos⁡17∘−sin⁡17∘cos⁡17∘+sin⁡17∘=cos⁡17∘(1−sin⁡17∘cos⁡17∘)cos⁡17∘(1+sin⁡17∘cos⁡17∘)⇒I=tan⁡45∘−tan⁡17∘1+tan⁡45∘⋅tan⁡17∘=tan⁡(45∘−17∘)=tan⁡28∘=tan⁡(90∘−62∘)=cot⁡62∘\begin{aligned} & \text{ Let } \mathrm{I}=\frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}=\frac{\cos 17^{{}^{\circ}}\left(1-\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}\right)}{\cos 17^{{}^{\circ}}\left(1+\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}\right)} \\ & \Rightarrow \mathrm{I}=\frac{\tan 45^{{}^{\circ}}-\tan 17^{{}^{\circ}}}{1+\tan 45^{{}^{\circ}} \cdot \tan 17^{{}^{\circ}}} \\ & =\tan \left(45^{{}^{\circ}}-17^{{}^{\circ}}\right)=\tan 28^{{}^{\circ}} \\ & =\tan \left(90^{{}^{\circ}}-62^{{}^{\circ}}\right)=\cot 62^{{}^{\circ}}\end{aligned}
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