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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If ^ 2 (x + y) + ^ 2 (x + y) = 1 - 2x - x^ 2 , then (where, n ∈ ℤ)

If tan⁡2(x+y)+cot⁡2(x+y)=1−2x−x2,then(where,n∈Z)\tan^{2} \left(x + y\right) + \cot^{2} \left(x + y\right) = 1 - 2x - x^{2}, then \left(where, n \in \mathbb{Z}\right)

Options

  1. Ax=1,y=nπ±π4−1x = 1, y = n\pi \pm \dfrac{\pi}{4} - 1
  2. Bx=−1,y=nπ±π4+1x = - 1, y = n\pi \pm \dfrac{\pi}{4} + 1
  3. Cx=−1,y=nπ2±π4−1x = - 1, y = \dfrac{n\pi}{2} \pm \dfrac{\pi}{4} - 1
  4. Dx=+1,y=nπ2±π4−1x = + 1, y = \dfrac{n\pi}{2} \pm \dfrac{\pi}{4} - 1

Correct answer

B. x=−1,y=nπ±π4+1x = - 1, y = n\pi \pm \dfrac{\pi}{4} + 1

Step-by-step solution

Left hand side is greater than or equal to 2.2.
⇒1−2x−x2≥2⇒ 1 - 2x - x^{2} \ge 2
⇒(x+1)2≤0⇒ {\left(x + 1\right)}^{2} \le 0
⇒x=−1⇒ x = - 1
Hence, the given equation is tan⁡2(y−1)+cot⁡2(y−1)=2\tan^{2} \left(y - 1\right) + \cot^{2} \left(y - 1\right) = 2
⇒tan⁡(y−1)=±1⇒ \tan \left(y - 1\right) = \pm 1
⇒y−1=nπ±π4⇒ y - 1 = n \pi \pm \dfrac{\pi}{4}
⇒y=(nπ±π4)+1⇒ y = \left(n \pi \pm \dfrac{\pi}{4}\right) + 1
So, x=−1x = - 1 and y=(nπ±π4)+1y = \left(n\pi \pm \dfrac{\pi}{4}\right) + 1
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