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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

If θ= 17^ ^ - 17^ ^ 17^ ^ + 17^ ^ , then what is the value of θ ?

If tan⁡θ=cos⁡17∘−sin⁡17∘cos⁡17∘+sin⁡17∘\tan \theta=\frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}, then what is the value of θ\theta ?

Options

  1. A0∘0^{{}^{\circ}}
  2. B28∘28^{{}^{\circ}}
  3. C38∘38^{{}^{\circ}}
  4. D52∘52^{{}^{\circ}}

Correct answer

B. 28∘28^{{}^{\circ}}

Step-by-step solution

tan⁡θ=cos⁡17∘−sin⁡17∘cos⁡17∘+sin⁡17∘=1−sin⁡17∘cos⁡17∘1+sin⁡17∘cos⁡17∘=1−tan⁡17∘1+tan⁡17∘\tan \theta=\frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}=\frac{1-\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}}{1+\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}}=\frac{1-\tan 17^{{}^{\circ}}}{1+\tan 17^{{}^{\circ}}}tan⁡θ=tan⁡45∘−tan⁡17∘1+tan⁡45∘⋅tan⁡17∘tan⁡θ=tan⁡(45∘−17∘)tan⁡θ=tan⁡(28∘)⇒θ=28∘\begin{aligned} & \tan \theta=\frac{\tan 45^{{}^{\circ}}-\tan 17^{{}^{\circ}}}{1+\tan 45^{{}^{\circ}} \cdot \tan 17^{{}^{\circ}}} \\ & \tan \theta=\tan \left(45^{{}^{\circ}}-17^{{}^{\circ}}\right) \\ & \tan \theta=\tan \left(28^{{}^{\circ}}\right) \Rightarrow \theta=28^{{}^{\circ}}\end{aligned}$$
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