Q90 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
$\lim\limits_{x \to \frac{\pi}{2}}\left(\dfrac{1 - \sin x}{\cos x}\right)$ is equal to
- A$-1$
- B$1$
- C$\dfrac{1}{2}$
- Correct answerD$0$
Solution
The given limit is of the form $\dfrac{0}{0}$ as $x \to \dfrac{\pi}{2}$.
Multiplying the numerator and the denominator by $1 + \sin x$:
$\lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{1 - \sin x}{\cos x} \times \dfrac{1 + \sin x}{1 + \sin x} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{1 - \sin^2 x}{\cos x (1 + \sin x)} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{\cos^2 x}{\cos x (1 + \sin x)} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{\cos x}{1 + \sin x} \right)$
Substituting $x = \dfrac{\pi}{2}$:
$= \dfrac{0}{1 + 1} = 0$
Answer: $0$
Multiplying the numerator and the denominator by $1 + \sin x$:
$\lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{1 - \sin x}{\cos x} \times \dfrac{1 + \sin x}{1 + \sin x} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{1 - \sin^2 x}{\cos x (1 + \sin x)} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{\cos^2 x}{\cos x (1 + \sin x)} \right)$
$= \lim_{x \to \dfrac{\pi}{2}} \left( \dfrac{\cos x}{1 + \sin x} \right)$
Substituting $x = \dfrac{\pi}{2}$:
$= \dfrac{0}{1 + 1} = 0$
Answer: $0$
Hence, the correct answer is option: 'D'.