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

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

If θ+ θ= 2 θ , then what is ( θ- θ) equal to?

If sin⁡θ+cos⁡θ=2cos⁡θ\sin \theta+\cos \theta=\sqrt{2} \cos \theta, then what is (cos⁡θ−sin⁡θ)(\cos \theta-\sin \theta) equal to?

Options

  1. A−2cos⁡θ-\sqrt{2} \cos \theta
  2. B−2sin⁡θ-\sqrt{2} \sin \theta
  3. C2sin⁡θ\sqrt{2} \sin \theta
  4. D2sin⁡θ2 \sin \theta

Correct answer

C. 2sin⁡θ\sqrt{2} \sin \theta

Step-by-step solution

sin⁡θ+cos⁡θ=2cos⁡θ\sin \theta+\cos \theta=\sqrt{2} \cos \theta...(1) Let cos⁡θ−sin⁡θ=P\cos \theta-\sin \theta=P...(2) (1)2+(2)2⇒sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ+sin⁡2θ−2sin⁡θcos⁡θ=2cos⁡2θ+p2⇒2=2cos⁡2θ+p2⇒p2=2(1−cos⁡2θ)=2sin⁡2θ⇒p=2sin⁡θ\begin{aligned} & (1)^2+(2)^2 \Rightarrow \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta+\cos ^2 \theta+\sin ^2 \theta- \\ & 2 \sin \theta \cos \theta=2 \cos ^2 \theta+\mathrm{p}^2 \\ & \Rightarrow 2=2 \cos ^2 \theta+\mathrm{p}^2 \\ & \Rightarrow \mathrm{p}^2=2\left(1-\cos ^2 \theta\right)=2 \sin ^2 \theta \\ & \Rightarrow p=\sqrt{2} \sin \theta\end{aligned}
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