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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2026 · Mathematics · Trigonometric Ratios & Identities

If x= θ- θ and y= θ+ θ , then which one of the following is correct...

If x=sec⁡θ−tan⁡θx=\sec\theta-\tan\theta and y=csc⁡θ+cot⁡θy=\csc\theta+\cot\theta, then which one of the following is correct ?

Options

  1. Ax+y−xy−1=0x+y-xy-1=0
  2. Bx−y+xy+1=0x-y+xy+1=0
  3. Cx+y+xy−1=0x+y+xy-1=0
  4. Dx−y+xy−1=0x-y+xy-1=0

Correct answer

B. x−y+xy+1=0x-y+xy+1=0

Step-by-step solution

Given x=sec⁡θ−tan⁡θ=1−sin⁡θcos⁡θx = \sec\theta - \tan\theta = \dfrac{1 - \sin\theta}{\cos\theta}

y=csc⁡θ+cot⁡θ=1+cos⁡θsin⁡θy = \csc\theta + \cot\theta = \dfrac{1 + \cos\theta}{\sin\theta}

Consider x(y+1)x(y+1):

x(y+1)=x(y+1) =\left(\dfrac{1 - \sin\theta}{\cos\theta}\right) \left(\dfrac{1 + \cos\theta}{\sin\theta} +1\right)<br><br><br><br>x(y+1) = \left(\dfrac{1 - \sin\theta}{\cos\theta} \right) \left(\dfrac{1 + \cos\theta + \sin\theta}{\sin\theta}\right)

x(y+1)=1+cos⁡θ+sin⁡θ−sin⁡θ−sin⁡θcos⁡θ−sin⁡2θsin⁡θcos⁡θx(y+1) = \dfrac{1 + \cos\theta + \sin\theta - \sin\theta - \sin\theta\cos\theta - \sin^2\theta}{\sin\theta\cos\theta}

x(y+1)=(1−sin⁡2θ)+cos⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θx(y+1) = \dfrac{(1 - \sin^2\theta) + \cos\theta - \sin\theta\cos\theta}{\sin\theta\cos\theta}

x(y+1)=cos⁡2θ+cos⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θx(y+1) = \dfrac{\cos^2\theta + \cos\theta - \sin\theta\cos\theta}{\sin\theta\cos\theta}

x(y+1)=cos⁡θ(cos⁡θ+1−sin⁡θ)sin⁡θcos⁡θx(y+1) = \dfrac{\cos\theta(\cos\theta + 1 - \sin\theta)}{\sin\theta\cos\theta}

x(y+1)=1+cos⁡θ−sin⁡θsin⁡θx(y+1) = \dfrac{1 + \cos\theta - \sin\theta}{\sin\theta}

x(y+1)=csc⁡θ+cot⁡θ−1x(y+1) = \csc\theta + \cot\theta -1

Substituting y=csc⁡θ+cot⁡θy = \csc\theta + \cot\theta:

x(y+1)=y−1x(y+1) = y - 1

xy+x−y+1=0xy + x - y + 1 = 0

Answer: x−y+xy+1=0x-y+xy+1=0
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