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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

If (45^ ^ +θ)=1+ 2 θ , where -(π)/(4) < θ < (π)/(4) , then what is...

If tan⁡(45∘+θ)=1+sin⁡2θ\tan \left(45^{{}^{\circ}}+\theta\right)=1+\sin 2 \theta, where −π4<θ<π4-\frac{\pi}{4} < \theta < \frac{\pi}{4}, then what is the value of cos⁡2θ\cos 2 \theta ?

Options

  1. A0
  2. B12\frac{1}{2}
  3. C1
  4. D2

Correct answer

C. 1

Step-by-step solution

tan⁡(45∘+θ)=1+sin⁡2θ⇒tan⁡45∘+tan⁡θ1−tan⁡45∘⋅tan⁡θ−1=2tan⁡θ1+tan⁡2θ\begin{aligned} & \tan \left(45^{{}^{\circ}}+\theta\right)=1+\sin 2 \theta \\ \Rightarrow & \frac{\tan 45^{{}^{\circ}}+\tan \theta}{1-\tan 45^{{}^{\circ}} \cdot \tan \theta}-1=\frac{2 \tan \theta}{1+\tan ^2 \theta}\end{aligned}⇒1+tan⁡θ1−tan⁡θ−1=2tan⁡θ1+tan⁡2θ⇒2tan⁡θ1−tan⁡θ=2tan⁡θ1+tan⁡2θ⇒2tan⁡θ(1+tan⁡2θ)=2tan⁡θ(1−tan⁡θ)⇒tan⁡2θ[1+tan⁡2θ−1+tan⁡θ]=0⇒tan⁡2θ(tan⁡θ+1)=0∴tan⁡θ=0 or  tan⁡θ=−1 (Not possible) \begin{aligned} & \Rightarrow \frac{1+\tan \theta}{1-\tan \theta}-1=\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & \Rightarrow \frac{2 \tan \theta}{1-\tan \theta}=\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & \Rightarrow 2 \tan \theta\left(1+\tan ^2 \theta\right)=2 \tan \theta(1-\tan \theta) \\ & \Rightarrow \tan ^2 \theta\left[1+\tan ^2 \theta-1+\tan \theta\right]=0 \\ & \Rightarrow \tan ^2 \theta(\tan \theta+1)=0 \\ & \therefore \tan \theta=0 \text{\text{ or } } \tan \theta=-1 \quad \text{ (Not possible) }\end{aligned}\Rightarrow \theta=0 \quad \left[\therefore \frac{-\pi}{4} < \theta < \frac{\pi}{4}\right] Now, \cos 2\theta=\cos \theta=11
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