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

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

What is (α+β)-2 α β+ (α-β) equal to ?

What is sin⁡(α+β)−2sin⁡αcos⁡β+sin⁡(α−β)\sin (\alpha+\beta)-2 \sin \alpha \cos \beta+\sin (\alpha-\beta) equal to ?

Options

  1. A0
  2. B2sin⁡α2 \sin \alpha
  3. C2sin⁡β2 \sin \beta
  4. Dsin⁡α+sin⁡β\sin \alpha+\sin \beta

Correct answer

A. 0

Step-by-step solution

sin⁡(α+β)−2sin⁡αcos⁡β+sin⁡(α−β)\sin (\alpha+\beta)-2 \sin \alpha \cos \beta+\sin (\alpha-\beta) by applying formula sin⁡C+sin⁡D=2sin⁡(C+D2)⋅cos⁡(C−D2)=2sin⁡α⋅cos⁡β−2sin⁡α⋅cos⁡β=0\begin{aligned} & \sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right) \\ & =2 \sin \alpha \cdot \cos \beta-2 \sin \alpha \cdot \cos \beta \\ & =0 \end{aligned}
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