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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If (θ-α), θ and (θ+α) are in AP, where α ≠ 1 , then what is the...

If sec⁡(θ−α),sec⁡θ\sec (\theta-\alpha), \sec \theta and sec⁡(θ+α)\sec (\theta+\alpha) are in AP, where cos⁡α≠1\cos\alpha \neq 1, then what is the value of sin⁡2θ+cos⁡α\sin ^2 \theta+\cos \alpha ?

Options

  1. A0
  2. B1
  3. C-1
  4. D1/2

Correct answer

A. 0

Step-by-step solution

2cos⁡θ=cos⁡(θ+α)+cos⁡(θ−α)cos⁡(θ+α)cos⁡(θ−α)=2cos⁡θ⋅cos⁡αcos⁡2θ−sin⁡2α⇒cos⁡2θcos⁡α=cos⁡2θ−sin⁡2α⇒sin⁡2α=cos⁡2θ(1−cos⁡α)⇒cos⁡2θ=sin⁡2α1−cos⁡α=1+cos⁡α⇒1−sin⁡2θ=1+cos⁡α⇒sin⁡2θ+cos⁡α=0\begin{aligned} & \frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2 \cos \theta \cdot \cos \alpha}{\cos ^2 \theta-\sin ^2 \alpha} \\ & \Rightarrow \cos ^2 \theta \cos \alpha=\cos ^2 \theta-\sin ^2 \alpha \\ & \Rightarrow \sin ^2 \alpha=\cos ^2 \theta(1-\cos \alpha) \\ & \Rightarrow \cos ^2 \theta=\frac{\sin ^2 \alpha}{1-\cos \alpha}=1+\cos \alpha \\ & \Rightarrow 1-\sin ^2 \theta=1+\cos \alpha \\ & \Rightarrow \sin ^2 \theta+\cos \alpha=0\end{aligned}
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