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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2023 · Paper 2 · Mathematics · Trigonometric Equations

General solution of 5 θ= 2 θ is

General solution of tan⁡5θ=cot⁡2θ\tan 5 \theta=\cot 2 \theta is

Options

  1. Aθ=nπ7+π14\theta=\frac{n \pi}{7}+\frac{\pi}{14}
  2. Bθ=nπ7+π5\theta=\frac{n \pi}{7}+\frac{\pi}{5}
  3. Cθ=nπ7+π2\theta=\frac{\mathrm{n} \pi}{7}+\frac{\pi}{2}
  4. Dθ=nπ7+π3\theta=\frac{\mathrm{n} \pi}{7}+\frac{\pi}{3}

Correct answer

A. θ=nπ7+π14\theta=\frac{n \pi}{7}+\frac{\pi}{14}

Step-by-step solution

We have tan⁡5θ=cot⁡2θ\tan 5 \theta=\cot 2 \theta
⇒tan⁡5θ=tan⁡(π2−2θ)…[∵tan⁡(π2−θ)=cot⁡θ]⇒5θ=nπ+π2−2θ⇒7θ=nπ+π2⇒θ=nπ7+π14.\begin{aligned} \\ & \Rightarrow \tan 5 \theta=\tan \left(\frac{\pi}{2}-2 \theta\right) \ldots \\ \\ & \left[\because \tan \left(\frac{\pi}{2}-\theta\right)=\cot \theta\right] \\ \\ & \Rightarrow 5 \theta=n \pi+\frac{\pi}{2}-2 \theta \Rightarrow 7 \theta=n \pi+\frac{\pi}{2} \\ \\ & \Rightarrow \quad \theta=\frac{n \pi}{7}+\frac{\pi}{14}. \\ \end{aligned}
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