Q178 / 270 · NDA 2026 (Phase 1) · Mathematics

If $x=\sec\theta-\tan\theta$ and $y=\csc\theta+\cot\theta$, then which one of the following is correct ?

  • A
    $x+y-xy-1=0$
  • Correct answer
    B
    $x-y+xy+1=0$
  • C
    $x+y+xy-1=0$
  • D
    $x-y+xy-1=0$

Solution

Given $x = \sec\theta - \tan\theta = \dfrac{1 - \sin\theta}{\cos\theta}$

$y = \csc\theta + \cot\theta = \dfrac{1 + \cos\theta}{\sin\theta}$

Consider $x(y+1)$:

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

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

$x(y+1) = \dfrac{1 + \cos\theta + \sin\theta - \sin\theta - \sin\theta\cos\theta - \sin^2\theta}{\sin\theta\cos\theta}$

$x(y+1) = \dfrac{(1 - \sin^2\theta) + \cos\theta - \sin\theta\cos\theta}{\sin\theta\cos\theta}$

$x(y+1) = \dfrac{\cos^2\theta + \cos\theta - \sin\theta\cos\theta}{\sin\theta\cos\theta}$

$x(y+1) = \dfrac{\cos\theta(\cos\theta + 1 - \sin\theta)}{\sin\theta\cos\theta}$

$x(y+1) = \dfrac{1 + \cos\theta - \sin\theta}{\sin\theta}$

$x(y+1) = \csc\theta + \cot\theta - 1$

Substituting $y = \csc\theta + \cot\theta$:

$x(y+1) = y - 1$

$xy + x - y + 1 = 0$

Answer: $x-y+xy+1=0$

Hence, the correct answer is option: 'B'.

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